TS Inter 2nd Year

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 4th Lesson Surface Chemistry Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 4th Lesson Surface Chemistry

Very Short Answer Questions (2 Marks)

Question 1.
What is an interface ? Give one example.
Answer:
The boundary that separate the two bulk phases is called surface or interface, e.g. Solid-liquid or solid/liquid.

Question 2.
What is adsorption ? Give one example.
Answer:
Accumulation of molecular species at the surface rather than in bulk of a solid or liquid is known as adsorption.
Eg : Aqueous solution of raw sugar when passed over animal charcoal becomes colourless due to the removal of colouring matter of sugar by adsorption on animal charcoal.

Question 3.
What is absorption? Give one example.
Answer:
Absorption is a process of distribution of the molecules of a substance uniformly in the bulk of another substance e.g: when a sponge or chalk piece is dipped in water, water is distributed throughout the sponge or chalk piece.

Question 4.
Distinguish between adsorption and absorption. Give one example.
Answer:

Eg : when a chalk piece is dipped in ink the colouring matter is retained only on the surface due to adsorption while the solvent of the ink goes deeper into the stick.

Question 5.
The moist air becomes dry in the presence of silica gel. Give reason for this.
Answer:
The moist air becomes dry in the presence of silica gel because water molecules get adsorbed on the surface of silica gel.

Question 6.
Methylene blue solution when shaken with animal charcoal gives a colourless filtrate on filtration. Give the reason.
Answer:
When animal charcoal is added to a solu¬tion of methylene blue solution and shaken the molecules of methylene blue are adsorbed on the surface of animal charcoal and are removed from solution. So the filtrate becomes colourless.

Question 7.
A small amount of silica gel and a small amount of anhydrous calcium chloride are placed separately in two corners of a vessel containing water vapour. What phenomena will occur ?
Answer:
Water vapour is absorbed by anhydrous calcium chloride but it is adsorbed on silica gel.

Question 8.
What is desorption ?
Answer:
The process of removal of an adsorbed substance from the surface is called desorption.

Question 9.
What is sorption ?
Answer:
If adsorption and absorption takes place simultaneously it is known as sorption.

Question 10.
Amongst adsorption, absorption which is a surface phenomena and why ?
Answer:
Adsorption is surface phenomenon. In this process molecular species accumulates at the surface rather than in the bulk.

Question 11.
What is the name given to the phenomenon when both adsorption and absorption take place together ?
Answer:
The name sorption is given to the phenomenon when both adsorption and absorption take place together.

Question 12.
Chalk stick dipped in an ink solution exhibits the following.
a) The surface of the stick retains the colour of the ink.
b) Breaking the chalk stick, it is found still white from inside.
Explain the above observations.
Answer:
a) The colouring matter is retained only on the surface due to adsorption. So the surface of the stick retains the colour.
b) Only the solvent of the ink goes deeper into the stick but not colouring matter. So it is still white inside.

Question 13.
What are the factors which influence the adsorption of a gas on a solid ?
Answer:

1. Nature of the gas
2. Nature of the adsorbent
3. Specific area of the solid
4. Pressure of the gas
5. Temperature
6. Activation of adsorbent

Question 14.
Why is adsorption always exothermic ?
Answer:
During adsorption there is always a decrease in residual forces of the surface i.e., there is decrease in surface energy which appears as heat. So adsorption is always exothermic.

Question 15.
Give the signs of AH and AS when ammonia gas gets adsorbed on charcoal.
Answer:
When ammonia gas is adsorbed on charcoal heat is liberated. So AH of adsorption is always negative (∆H = -Ve). When ammonia gas is adsorbed on charcoal the freedom of its molecules become restricted. So entropy decreases (∆S = -Ve).

Question 16.
How many types of adsorption are known? What are they?
Answer:
There are mainly two types of adsorption.
a) Physical adsorption or physisorption.
b) Chemical adsorption or chemisorption.

Question 17.
What types of forces are involved in physisorption of a gas on solid ?
Answer:
Only weak van der Waal’s forces are involved in physisorption of a gas on solid.

Question 18.
What type of interaction occuring between gas molecules and a solid surface is responsible for chemisorption of the gas on solid?
Answer:
Chemical bonds occur between gas molecules and a solid surface during chemisorption of the gas on solid. The chemical bonds may be ionic or covalent in nature.

Question 19.
Why chemisorption is called activated adsorption ?
Answer:
The chemical adsorption involves a high energy of activation. So it is called activated adsorption.

Question 20.
What is the difference between physisorption and chemisorption ?
Answer:
Physisorption is due to weak van der Waal’s forces between the gas molecules and the solid surface. Chemisorption is due to chemical bonds such as covalent or ionic between gas molecules and solid surface.

Question 21.
Out of physisorption and chemisorption, which can be reversed ?
Answer:
Physisorption can be reversed. By increasing the pressure more gas is adsorbed and by decreasing the pressure the adsorbed gas is removed. By decreasing the temperature more gas is adsorbed while by increasing the temperature the adsorbed gas is removed.

Question 22.
How is adsorption of a gas related to its critical temperature ? ,
Answer:
Easily liquefiable gases which have high critical temperatures are readily adsorbed since the van der Waal’s forces are stronger near the critical temperatures.

Question 23.
The critical temperature of SO₂ is 630K and that of CH₄ is 190K. Which one is adsorbed easily on actived charcoal ?
Answer:
Activated charcoal adsorbs more SO₂ since it has more critical temperature than methane.

Question 24.
Easily liquifiable gases are readily adsorbed on solids. Why ?
Answer:
Easily liquifiable gases are readily adsorbed on solids because van der Waal’s forces are stronger near the critical temperature.

Question 25.
Amongst SO₂, H₂ which will be adsorbed more readily on the surface of charcoal and why ?
Answer:
SO₂ is adsorbed more readily on the surface of charcoal compared to H₂ because its critical temperature is more, as van der Waal’s forces are stronger near critical temperature.

Question 26.
Compare the enthalpy of adsorption for physisorption and chemisorption.
Answer:
The enthalpy of adsorption is low in physisorption (20 – 40 kJ mol^-1) but in chemisorption it is high (80 – 240 kJ mol^-1).

Question 27.
What is the magnitude of enthalpy of physical adsorption ? Give reason for this magnitude.
Answer:
The magnitude of enthalpy of physical adsorption is low i.e., about 20 -40 kJ mol^-1. This is because the attraction between gas molecules and solid surface is only due to weak van der Waal’s forces.

Question 28.
What is the magnitude of enthalpy of chemisorption ? Give reason for this magnitude.
Answer:
The enthalpy of chemisorption is high 80 – 240 KJ mol^-1 as it involves chemical bond formation.

Question 29.
Give any two applications of adsorption.
Answer:

1. Gas masks containing activated charcoal or mixture of adsorbents used by coal miners adsorb poisonous gases during breathing.
2. Due to the difference in degree of adsorption of gases by charcoal a mixture of noble gases can be separated by adsorption on coconut charcoal at different temperatures.

Question 30.
Why physisorption suffers from lack of specificity ?
Answer:
Since van der Waal’s forces are universal, a given surface of an adsorbent does not show any preference for a particular gas. So physisorption suffers from lack of specificity.

Question 31.
What is an adsorption isotherm ? Write the equation of Freundlich adsorption isotherm.
Answer:
The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be shown as a curve termed as adsorption isotherm.
Freundlich adsorption isotherm is
\(\frac{\mathrm{x}}{\mathrm{m}}\) = k . p^1/n (n > 1) m
where x is the mass of the gas adsorbed on mass ‘m’ of the adsorbent at pressure p. k and n are constants.

Question 32.
In the Freundlich adsorption isotherm mention the conditions under which following graph will be true ?

Answer:
When \(\frac{\mathrm{1}}{\mathrm{n}}\) = 0 in Freundlich adsorption isontherm, \(\frac{\mathrm{x}}{\mathrm{m}}\) = constant, the adsorption is m independent of pressure.

Question 33.
What role does adsorption play in heterogeneous catalysis ?
Answer:
Adsorption of reactants on solid surface of the catalysts increase the rate of reaction. There are many gaseous reactions of industrial importance involving use of solid catalysts.

Question 34.
What is the role of MnO₂ in the preparation of O₂ from KClO₃ ?
Answer:
In the preparation of O₂ from KClO₃, MnO₂ acts as catalyst and the decomposition of KClO₃ takes place at considerably low temperatures.

Question 35.
Define “promoters” and “poisons” in the phenomenon of catalysis ?
Answer:
Promoters are the substances that enhances the activity of catalyst. Poisons are the substances which decrease the catalytic activity of a catalyst.

Question 36.
What is homogeneous catalysis ? How is different from heterogeneous catalysis ?
Answer:
If the reactants and the catalysts are in the same phase, the process is said to be homogeneous catalysis. In heterogeneous catalysis the reactants and catalysts are in different phases.

Question 37.
Give two examples for homogeneous catalytic reactions.
Answer:
1) Oxidation of SO₂ to SO₃ with O₂ in the presence of oxides of N₂ as catalyst in the lead chamber process.
2SO₂(g) + O₂ (g) > SO₃ (g)
Reactants and catalysts are gases

2) Acid catalysed hydrolysis of ester
CH₃ COOC₂H₅(l) + H₂O CH₃COOH(aq) + C₂H₅ – OH(aq)
Reactants and catalysts are liquids

Question 38.
Give two examples for heterogeneous catalysis.
Answer:
1) Oxidation of sulphur dioxide into sulphur trioxide in the presence of Pt
2 SO₂(g) + O₂(g) 2SO₃
Reactants are gases while the catalyst Pt is solid.

2) Manufacture of ammonia by Haber’s process involves the combination of N₂ and H₂ in the presence of iron powder.
N₂(g) + 3H₂(g) 2NH₃ (g)
Reactants are gases catalyst is solid.

Question 39.
Give two examples which indicate the selectivity of heterogeneous catalysis.
Answer:
Selectivity of a catalyst is its ability to direct a reaction to from specific products.

Question 40.
Why zeolites are treated as shape selective catalysts ?
Answer:
Zeolites are good shape selective catalysts because of their honeycomb like structures.
The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules as well as upon the pores and cavities of the zeolites.

Question 41.
Which zeolite catalyst is used to convert alcohols directly into gasoline ?
Answer:
ZSM-5 converts alcohols directly into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons.

Question 42.
What are enzymes ?What is their role in human body ?
Answer:
Enzymes are biochemical catalysts and the phenomenon of catalysis is called biochemical catalysis. These are complex nitrogenous organic compounds. They are protein molecules of higher molecular mass and form collidal solutions in water. Enzymes catalyse the life process in human body.

Question 43.
Can catalyst increase the yield of reaction?
Answer:
No. A catalyst will increase the rates of forward and backward reactions equally. So the yield of reaction do not change.

Question 44.
Name any two enzyme catalysed reactions. Give the reasons.
Answer:
i) Inversion of sugar: The enzyme invertase converts cane sugar into glucose and fructose.

ii) Conversion of glucose into ethyl alcohol:
The enzyme zymase converts glucose into ethyl alcohol and carbon dioxide.

Question 45.
Name the enzymes obtained from soya-beans source.
Answer:
Urease which converts urea into ammonia and carbon dioxide.

Question 46.
Name the used in
a) Decomposition of urea into ammonia.
b) Conversion of proteins into peptides in stomach.
Answer:
a) Decomposition of urea of into ammonia is catalysed by urease.
NH₂CONH₂ (aq) → 2NH₃(g) + CO₂(g)
b) Pepsin converts proteins into peptides in stomach.

Question 47.
What enzymes are obtained from yeast ?
Answer:

1. Invertase which converts sugar into glucose and fructose,
2. Zymase which converts glucose into ethyl alcohol and carbondioxide.

Question 48.
At what ranges of temperature and pH, enzymes are active ?
Answer:
The optimum temperature range for enzymatic activity is 298 – 310K.
The pH range for enzymatic activity is 5-7.

Question 49.
Represent diagramatically the mechanism of enzyme catalysis.
Answer:

Question 50.
Name any two industrially important heterogeneous catalytic reactions mentioning the catalysts used.
Answer:

1. In Haber’s process for the manufacture of ammonia, catalyst is iron powder mixed with molybdenum as promoter.
2. In Contact process for the manufacture of sulphuric acid, catalyst is platinised asbestos or vanadium pentoxide.

Question 51.
What is a colloidal solution ? How is it different from a true solution with respect to dispersed particle size and homogeneity ?
Answer:
A colloidal solution is a heterogeneous sys-tem in which one substance is dispersed (dispersed phase) as large particles in another substance (dispersion medium) Colloidal particles are larger than simple molecules of a true solution. Their size range between 1 and 1000 nm (10^-9 to 10^-6m).

Question 52.
Name the dispersed phase and dispersion medium in the following colloidal system i) fog ii) smoke iii) milk.
Answer:

1. Fog : Dispersed phase : Liquid
   Dispersed medium : Gas
2. Smoke : Dispersed phase : Solid
   Dispersed medium : Gas
3. Milk : Dispersed phase : Fat
   Dispersed medium : Water

Question 53.
What are lyophilic and lyophobic sols ? Give one example for each type.
Answer:
Colloids in which there is affinity between dispersed phase and dispersion medium are called lyophilic sols.
Ex : Starch sol, Gelatin. ,
Colloids in which there is very little affinity between the dispersion medium, and dispersed phase are called lyophobic sols.
Ex : Smoke, Gold sol.

Question 54.
Explain the terms with suitable examples.
i) aerosol
ii) hydrosol
Answer:
i) Aerosol is a colloidal system in which dispersed phase is liquid and dispersion medium is gas. e.g : fog, smoke.
ii) Hydrosol is the colloidal system that consists of water as dispersion medium. e.g : Starch in water, milk.

Question 55.
Explain why lyophilic colloids are relatively more stable than lyophobic colloids.
Answer:
Lyophilic colloids are more stable than lyophobic colloids because lyophilic colloids are extensively solvated. The colloidal particles are covered by a sheath of the dispersion medium in which they are dispersed.

Question 56.
Give two examples of colloidal solutions of liquids dispersed in solid. What is the name given to colloidal solution ?
Answer:

1. Butter
2. Cheese
   Liquids dispersed in solid type colloids are named as gels.

Question 57.
What is the difference between multimolecular and macromolecular colloids ? Give one example for each.
Answer:
In multimolecular colloids large number of atoms or small molecules of the dispersed phase aggregate together to form species in the colloidal range. e.g.: Gold sol, sulphur sol.
In macromolecular colloids the size of individual macromolecules are in the colloidal range.
e.g.: Starch, proteins etc.

Question 58.
What are micelles ? Give one example.
Answer:
The substances which behave as electrolytes at low concentration but at higher concentrations exhibit colloidal behaviour due to formation of aggregates are called micelles and the colloids formed are called associated colloids. They contain a hydrophobic long hydrocarbon chain as one end and a hydrophilic polar group as another end. e.g.: Sodium stearate R COO^– Na⁺.

Question 59.
How do micelles differ from a normal collodial solutions ?
Answer:
Micelles contain a long hydrocarbon part which is hydrophobic tail and a polar hydrophilic head. These are absent in normal colloidal solution.
eg. Sodium stearate C₁₇H^–₃₅COO^–Na⁺ contain long hydrocarbon part stearate radical (C₁₇H₃₅^–) is hydrophobic tail while COO^– part is hydrophilic head.

Question 60.
Give two examples of associated colloids.
Answer:

1. Soaps dissolved in water
2. Detergents dissolved in water.

Question 61.
Can the same substance act both as colloid and crystalloid ?
Answer:
Yes. The same substance can act both as colloid and crystalloid.
e.g.: Common salt
(NaCl) a typical crystalloid in an aqueous solution behave as a colloid in the benzene medium.

Question 62.
Give two examples of lyophobic sols.
Answer:

1. Metal colloids like gold sol.
2. Collids of metal sulphides such as AS₂S₃.

Question 63.
Give examples of colloidal system of
i) Liquid in solid
ii) gas in solid.
Answer:
i) Example for liquid in solid is cheese or butter.
ii) Example for gas in liquid is froth or soap lather.

Question 64.
What type of substances form lyophobic sols?
Answer:
Substances like metals, and their sulphides form lyophobic sols.

Question 65.
What is Critical Micelle Concentration (CMC) and Kraft temperature ?
Answer:
The formation of micelle takes place above certain temperature called Kraft temperature and above a particular concentration called Critical Micelle Concentration.

Question 66.
Why lyophobic colloids are called irreversible colloids ?
Answer:
Lyophobic sols are readily precipitated on the addition of small amounts of electrolytes or by heating or by shaking. The precipitates does not give back the collodial sol by simple addition of the dispersion medium to it. So they are called irreversible colloids.

Question 67.
How a colloidal sol of arsenous sulphide is prepared ?
Answer:
Arsenous sulphide sol can be prepared by ; double decomposition method of arsenous i oxide and hydrogen sulphide.
AS₂O₃ + 3H₂ S → AS₂S₃ (sol) + 3H₂O

Question 68.
What is peptization ?
Answer:
Peptization is a process of converting a precipitate into colloidal sol by shaking it with the dispersion medium in the presence of a small amount of electrolyte. The electrolyte used for this purpose is called peptizing agent.

Question 69.
What is dialysis ? How is dialysis can be made feat ?
Answer:
Dialysis is a process of removing of dissolved substance from a collodial solution using a suitable membrane. Dialysis can be made fast by applying emf if the dissolved substance in impure colloidal solution is an electrolyte i.e., electrodialysis.

Question 70.
What is collodion solution ?
Answer:
Collodion is a 4% solution of nitro – cellulose in a mixture of alcohol and ether.

Question 71.
How an ultrafilter paper is prepared from ordinary filter paper ?
Answer:
Ultra filter paper is prepared by soaking the filter paper in collodion solution hardening by formaldehyde and then finally drying it.

Question 72.
What is Tyndall effect ?
Answer:
Colloidal particles scatter light in all directions in space. The scattering of light illuminates the path of beam in the colloidal dispersion. This is known as Tyndall effect.

Question 73.
Under what conditions is Tyndall effect observed ?
Answer:
Tyndall effect is observed only when the following two conditions are satisfied.

1. The diameter of the dispersed particles is not much smaller than the wavelength of the light used.
2. The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.

Question 74.
Can Tyndall effect be used to distinguish between a collodial solution and a true solution ? Explain.
Answer:
Tyndall effect is used to distinguish between a colloidal solution and a true solution. When an intense beam of light is focussed on the collodial solution contained in a glass vessel, the focus of light can be observed with a microscope kept at right angles to the beam. Individual colloidal particles appear as bright stars.

Question 75.
Sky appears blue in colour. Explain.
Answer:
Dust particles along with water vapour suspended in air, scatter blue light which reaches our eyes and hence the sky looks blue.

Question 76.
What is Brojvnian movement ?
Answer:
The zig-zag motion of colloidal particles all over the field of view in a collodial solution is called Brownian movement.

Question 77.
What is the main cause for charge on a colloidal solution ?
Answer:
Preferential adsorption of ions is the main cause for charge on a colloidal particle. The sol particles acquire positive or negative charge by preferential adsorption of positive or negative ions.

Question 78.
What is electrokinetic potenital or zeta potential ?
Answer:
The potential difference between the fixed layers and the diffused layer of opposite charge in a colloidal solution is called the electrokinetic potential or zeta potential.

Question 79.
Write the formula of positively charged and negatively charged hydrated ferric oxide collodial solutions.
Answer:
If FeCl₃ solution is added to hot water a positively charged sol of hydrated ferric oxide is formed due to adsorption of Fe³⁺ ions. However when ferric chloride solution is added to NaOH solution taken in excess, a negatively charged sol is obtained with adsorption of OH^– ions.
Fe₂O₃ . xH₂O / Fe ³⁺
Positively charged
Fe₂O₃ . xH₂O /OH^–
Negatively charged

Question 80.
Give the order of coagulating power of Cl^–, SO₄^2-, PO₄^3- in the coagulation of positive sols.
Answer:
PO₄^3- > SO₄^2- > Cl^–
More the charge on anion more is the coagulating power.

Question 81.
Amongst Na⁺, Ba²⁺, Al³⁺ which coagulates negative sol readily and why ?
Answer:
According to Hardy – Schulze rule more the charge on ion more is the coagulating power of the oppositely charged colloid. Since Al carry more charge coagulates negative sol readily.

Question 82.
A colloidal solution of AgI is positively charged when prepared from a solution containing excess of Ag⁺ ions and negatively charged when prepared solution containing excess of I^– ions. Explain.
Answer:
A collodial particle adsorb the ion from solution which is common to it. When AgI colloid is prepared from a solution containing excess Ag⁺ ions, the colloidal particle adsorb Ag⁺ ion and thus get positive charge. If Agl colloid is prepared from solution containing excess I^– ions, the colloidal particle adsorb I^– ions and thus get negative charge.

Question 83.
What is electrophoresis ?
Answer:
The movement of colloidal particles under an applied emf is called electrophoresis.

Question 84.
What is electro osmosis ?
Answer:
If the movement of colloidal particles is arrested by some suitable means, the dispersion medium moves in opposite direction under the applied emf. This phenomenon is called electro osmosis.

Question 85.
What is coagulation ?
Answer:
The process of settling down of colloidal particles is called coagulation or precipitation or flocculation of the sol.

Question 86.
Define flocculation value.
Answer:
The minimum concentration of an electrolyte in millimoles per litre required to cause coagulation of a sol in two hours is called flocculation value.

Question 87.
State Hardy – Schulze rule.
Answer:
Greater the valence (charge) of the coagu-lating ion added greater is its power to cause coagulation. This is known as Hardy-Schulze rule.

Question 88.
Coagulation takes place when sodium chloride solution is added to a colloidal solution of hydrated ferric oxide. Explain.
Answer:
When sodium chloride solution added to a colloidal solution of hydrated ferric oxide, the Na⁺ ions neutralize the negative charge on colloidal particles then colloidal particles come close, combine forming bigger particles and thus coagulated.

Question 89.
How are lyophobic solutions protected from phenomenon of coagulation ?
Answer:
Lyophobic colloids are protected from coagulation by adding lyophilic colloids to lyophobic colloid. The lyophilic colloid form a protective layer around lyophobic particles and thus protect the lyophobic colloid from the action of electrolytes.

Question 90.
What is protective colloid ?
Answer:
When lyophilic colloid is added to lyophobic colloid, the lyophilic colloid form a protective layer around lyophobic colloid and thus protect the lyophobic colloid from the action of electrolytes. So the lyophilic colloids are called protective colloids.

Question 91.
What is an emulsion ? Give two examples.
Answer:
Emulsions are liqud – liquid colloidal systems in which finely divided droplets of one liquid dispersed in another. If a mixture of two immiscible or partially miscible liquids is shaken a coarse dispersion of one liquid in the other called enjulsion is obtained.
Ex : Milk and Vanishing cream.

Question 92.
How emulsions are classified ? Give one example for each type of emulsion. [Mar. 2018 . TS]
Answer:
Emulsions are two types.

1. Oil dispersed in water or oil/water (o/w) type eg. Milk, Vanishing cream.
2. Water dispersed in oil (water / oil) (w/o) type. eg. Butter, cream.

Question 93.
What is an emulsifying agent ?
Answer:
For stablisation of an emulsion, the third component added to emulsion is called emulsifying agent. Eg (Proteins and soaps are emulsifying agent for o/w emulsions.

Question 94.
What is demulsification ? Name two demulsifiers.
Answer:
Separation of an emulsion into consituent liquids is called demulsification. Demulsification can be carried by heating, freezing, centrifusing etc.

Question 95.
How is artifical rain produced ?
Answer:
Artificial rain can be produced by throwing electrified sand or spraying a sol carrying charge opposite to the one on clouds from an aeroplane.

Question 96.
Bleeding from fresh cat can be stopped by applying alum. Give reasons.
Answer:
Blood is a colloidal solution of alluminoid substances. The styptic action of alum is due to coagulation of blood forming a clot which stops further bleeding.

Question 97.
Deltas are formed at the points where river enters the sea. Why ?
Answer:
River water is a colloidal solution of clay. Sea water contains a number of electrolytes. When river water meets the sea water, the electrolytes of sea water coagulates the colloidal solution of clay resulting in its deposition of clay with formation of delta.

Question 98.
Name any two applications of colloidal solutions.
Answer:

1. Purification of drinking water : Water from natural sources contain collodial impurities. When alum is added the colloidal particles in water coagulates and the water becomes fit for drinking.
2. Several medicines are used as colloi-dal solutions : Colloidal medicines are more effective due to their large surface area and are therefore easily assimilated.

Question 99.
How can aerial pollution by colloidal particles of smoke be prevented ? Explain.
Answer:
Smoke is a colloidal solution of solid particles such as carbon, arsenic compounds, dust etc. The smoke is made to pass through a precipitator called Cottrell precipitator. Before coming out of chimney the precipitator contain plates carrying opposite charge to that carried by smoke particles. The particles on coming in contact with these plates lose their and get precipitated.

Question 100.
Alum is used to purify water obtained from natural soruces. Explain.
Answer:
Water from natural sources contain colloidal impurities. When alum is added, the colloidal particles in water coagulates and the water becomes fit for drinking.

Question 101.
Why medicines are more effective in colloidal state ?
Answer:
Medicines in colloidal state are more effective due to their large surface area and are therefore easily assimilated.

Question 102.
How rubber is obtained from latex ?
Answer:
Plant latex is a colloidal solution of rubber particles which are negatively charged. Rubber is obtained from latex by coagulation on adding electrolytes.

Question 103.
Name the type of emulsion to which milk belongs.
Answer:
Milk is an emulsion of oil dispersed in water (o/w) type.

Short Answer Questions (4 Marks)

Question 104.
What is adsorption? Discuss the mechanism of adsorption of gases on solids.
Answer:
The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is known as adsorption.

Mechanism of adsorption :
Adsorption arises due to the fact that the surface particles of the adsorbent are not in the same environment as the particles inside the bulk. The particle inside the bulk is surrounded by other particles in all directions and all the forces acting between the particles are mutually balanced.

But the particle on the surface have no particles of its own above it and contain only below to it. So these unbalanced residual attractive forces are responsible for attracting the adsorbate molecules on its surface.

Question 105.
What are different types of adsorption ? Give any four differences between characteristics of these different types. [TS Mar. 19; (AP & TS 15)]
Answer:
Adsorption is two types, i) Physical adsor-ption and 2) Chemical adsorption.

Homogeneous catalysis: When the rea­ctants and the catalyst are in the same phase i.e., either liquid or gas the process is said to be homogeneous catalysis.

Question 106.
What do you understand by the terms given below ?
a) Absorption
b) Adsorption
c) Adsorbent and adsorbate
Answer:
a) Absorption: Uniform distribution of the substance throughout the bulk of a solid is absorption.
b) Adsorption: Accumulation of molecular species at the surface rather than in the bulk of solid or liquid is known as adsorption.
c) The substance which concentrates or accumulates on the surface is termed as adsorbate and the material on the surface of which the adsorption takes place is called adsorbent.

Question 107.
Adsorption of a gas on the surface of solid is generally accompanied by decrease in entropy. Still it is a spontaneous process. Explain.
Answer:
During adsorption there is always decrease in surface energy which appears as heat. So adsorption is always exothermic and ∆H of adsorption is always negative. But during adsorption entropy of the gas decreases due to decrease in the movement of gas molecules.

Thus entropy of adsorption AS is also negative. For the process of adsorption ∆G must be negative at a constant temperature on the basis of equation ∆G = ∆H – T∆S. In the equation ∆G = ∆H – T∆S, ∆G can be negative if ∆H has sufficiently more negative value and -T∆S is positive. The combination of these two factors makes ∆G negative. So adsorption process becomes spontaneous.

As the adsorption proceeds ∆H become less negative and becomes equal to T∆S and ∆G become zero. At this state equilibrium is attained.

Question 108.
How can the constants k and n of the Freundlich adsorption equation be calculated ?
Answer:
Freundlich adsorption isotherm is
\(\frac{\mathrm{x}}{\mathrm{m}}\) – k . p^1/n (n > 1) ……………… (1)
Where x is the mass of gas adsorbed
m is the mass of adsorbent
p is the pressure
Taking logarithms of equation (1)
log \(\frac{\mathrm{x}}{\mathrm{m}}[/latex = logk + [latex]\frac{\mathrm{1}}{\mathrm{n}}[/latex log p

A graph is plotted by taking [latex]\frac{\mathrm{x}}{\mathrm{m}}\) on Y – axis and log p on X – axis. This gives a straight line the slope of the straight line gives the value of \(\frac{\mathrm{1}}{\mathrm{n}}\) . The intercept on the Y – axis n gives the value of log k.

Question 109.
How does the extent of adsorption depend upon
a) Increasing the surface area per unit mass of adsorbent
b) Increasing temperature of the system
c) Increasing pressure of the gas.
Answer:
a) The surface area available for adsorption per unit mass of adsorbent is known as specific area. Greater the specific area of the solid, greater would be its adsorbent power. Because of this reason porous or finely divided forms of adsorbents adsorb more extensively.

b) The adsorption at a surface initially increases till a saturation point is achieved. At this juncture an equilibrium is established.
Adsorption ⇌ Desorption ; ∆H = +Ve Since adsorption is accompanied by evolution of heat, according to Le-Chateliers principle the magnitude of adsorption will decrease with rise in temperature.

c) The adsorption of gas generally increases the increase of pressure at constant temperature. Adsorption of gas on solid resuits in the decrease of pressure, therefore according to Le-Chatelier’s principle increase in pressure increase the extent of adsorption.

Question 110.
What is catalysis? How is catalysis classified ? Give two examples for each type of catalysis [AP 16; TS 15; IPE ’14]
Answer:
Substances which accelerate the rate of a slow chemical reaction but whose chemical nature and mass remain unchanged after the reaction are known as catalysts and the phenomenon is known as catalysis.
Catalysis is mainly two types.
i) Homogeneous catalysis: When the reactants and the catalyst are in the same phase i.e., either liquid or gas the process is said to be homogeneous catalysis.
e-g

Reactants SO₂ and O₂ and the catalyst NO are all gases and in the same phase.

ii) CH₃COOCH₃ (l) + H₂O CH₃COOH (aq) + C₂H₅OH (aq)
The reactants and catalysts are in the same liquid phase.

2) Heterogeneous catalysis: The catalytic process in which the reactants and the catalyst are in different phases is known as heterogeneous catalysis.

The reactants are in gaseous state while that catalyst is solid.

Reactants are gases, catalyst is solid.

Question 111.
Discuss the mechanism involved in adsorption theory of heterogeneous catalysis.
Answer:
As per old theory the reactant molecules in the gaseous state or in solutions are adsorbed on the surface of the solid catalyst. The increase in concentration of the reactants on the surface increase the rate of reaction. The heat liberated during adsorption is also utilised in increasing the rate of reaction. According to modern theory,

1. Reactant molecules diffuse on to the surface of catalyst.
2. The reactant molecules adsorb on the surface of the catalyst.
3. Chemical reaction takes place between adsorbed reactant molecules through formation of an intermediate.
4. The product molecules adsorb from the catalyst surface making the surface available for more reaction to occur.
5. The product molecules diffuse away from the surface of the catalyst for the adsorption of fresh reactant molecules.

Question 112.
Discuss some features of catalysis by zeolites.
Answer:
The catalytic reactions which depend on the structure of the pores of catalyst and the size of the molecules of reactants and products is called shape – selective catalysis.

Zeolites are good shape- selective catalysts, because of their honey comb like structure. Zeolites are microporous alumino silicates with three dimensional network of silicates. The reactions taking place in zeolites are shape – selective catalytic reactions.

Zeolites are widely used as catalysts in photo chemical industries for cracking and isomerisation of hydrocarbons. Eg: ZSM – 5 converts alcohols directly into petrol by dehydrating them.

Question 113.
Give brief account of mechanism of enzyme catalysis with suitable diagrams.
Answer:

Colloidal enzyme particles contain large number of cavities which have characteristic shape. They possess active groups such as -NH₂, -COOH, -SH, -OH etc. These are active centres on the surface of the enzyme particles. The molecules of the reactant which have a complimentary shape that can fit into these cavities just like a key fit into a lock.

The active groups combine with reactant molecule forming an activated complex which then decompose into products.

The enzyme catalysed reactions may be considered to proceed in two steps.

1. Binding of substrate to enzyme to form an activated complex (ES#).
   E + S → ES#
2. Activated complex decomposes to give product.
   ES → E + P

Question 114.
Discuss the factors that influence the catalytic activity of enzymes.
Answer:
The catalytic activity of enzymes will be influenced by
1) Temperature: The rate of enzyme catalysed reaction is maximum at a definite temperature called the optimum temperature. The optimum range is 298 – 310K. On either side of the optimum temperature the enzyme activity decreases.

2) pH : The rate of an enzyme – catalysed reaction is maximum at a particular pH called optimum pH which lies between 5-7.

3) Activators and Co-enzymes: The enzymatic activity is increased in the presence of certain substances known as Co-enzymes e.g. If a small non-protein vitamin is present along with an enzyme the catalytic activity is enhanced considerably.
Activators are generally metal ions Na⁺, Mn²⁺, CO²⁺, Cu²⁺ etc. When these are bonded to enzyme molecules their catalytic activity increases.
e.g. Catalytic activity of amylase combined with Na+ is very high.

4) Inhibitors or poisons : The substance which decrease the catalytic activity of enzymes are called inhibitors or poisons. These interact with the active functional groups and reduce the catalytic activity of the enzyme.

Question 115.
Name any six enzyme catalysed reactions.
Answer:

1. Inversion of cane sugar into glucose and fructose in the presence of enzyme invertase.
   
2. Conversion of glucose into ethyl alcohol in the presence of zymase.
   
3. Conversion of starch into maltose in the presence of diastase.
   
4. Conversion of maltose into glucose in the presence of maltase.
   
5. Conversion of proteins into peptides in intestine in the presence of enzyme trypsin.
6. Conversion of milk into curd in the presence of enzyme lacto bacilli.

Question 116.
What do you mean by activity and selectivity of catalysts ?
Answer:
Activity :
The ability of a catalyst in increasing the rate of reaction is defined as its activity. The activity of the catalyst depends upon the strength of chemisorption. When the reactant molecules are adsorbed on the surface of catalyst they become reactive.

Selectivity:
The selectivity of a catalyst is its ability to direct a reaction to form specific products. Eg. Different products are formed from H₂ and CO by using different catalysts.

These reactions indicate that a given substance can act as catalyst only in a particular reaction and not in all reactions.

Question 117.
How are colloids classified on the basis of physical states of components ?
Answer:
Depending on the physical states of components the colloids are classified into eight types

Question 118.
How are colloids classified on the basis of nature of the dispersion medium ?
Answer:
Dispersion medium may be a solid, liquid or gas. Thus depending on the nature of dispersion medium the colloids are three types.

1. Dispersion medium is solid: Dispersion phase may be solid or liquid or gas.
2. Dispersion medium is liquid: Dispersion phase may be solid or liquid of gas.
3. Dispersion medium is gas: Dispersion phase may be solid or liquid.
4. Solid in liquids are called sols, liquid in liquid are called emulsions, liquid in solids are called gels. If the dispersion medium is water, the sol is called aquasol or hydrosol and if the dispersion medium is alcohol it is called alcohol and so on.

Question 119.
How are colloids classified on the basis of interaction between dispersed phase and dispersion medium ?
Answer:
Depending upon the nature of interaction between the dispersed phase and dispersion medium the colloidal sols are classified into two types.
1) Lyophilic colloids:
If there is attraction between the dispersion medium and dispersed phase, the colloids are called as lyophilic colloids. Lyophilic means solvent loving or solvent attracting. These can be prepared by direct mixing and can be separated from one another by evaporating. These sols are reversible, so called reversible sols. These are stable and cannot be easily coagulated.

2) Lyophobic colloids :
If there is repulsion between dispersion medium and dispersed phase, the colloids are called lyophobic colloids. Lyophobic means solvent hating. This type of colloids cannot be prepared simply by mixing. They have to be prepared by special methods. These sols are readily precipitated by the addition of small amounts of electrolytes or by heating. The precipitate does not give back the colloidal sol by simple addition of the dispersion medium to it. So these are known as irreversible sols. Lyophobic sols need stabilising agents for their preservation.
If water is the dispersion medium these are called hydrophilic and hydrophobic colloids.

Question 120.
What Is the difference between a colloidal sol, gel, emulsion and a foam ?
Answer:
When the dispersed phase is solid and the dispersion medium is liquid the colloid is called sol. Ex: Paints.
If the dispersed phase is liquid and the dispersion medium is solid, the colloid is called gel. Ex: Cheese, butter.
If one liquid is dispersed in another liquid, the colloid is called emulsion. Ex: Milk
If the dispersed phase is gas and dis-persed medium is liquid, the colloid is called foam. Ex : Froth, soap lather.

Question 121.
What are lyophilic and lyophobic sols ? Compare two terms interms of stability and reversibility.
Answer:
Depending upon the nature of interaction between the dispersed phase and dispersion medium the colloidal sols are classified into two types.
1) Lyophilic colloids:
If there is attraction between‘the dispersion medium and dispersed phase, the colloids are called as lyophilic colloids. Lyophilic means solvent loving or solvent attracting. These can be prepared by direct mixing and can be separated from one another by evaporating. These sols are reversible, so called reversible sols. These are stable and cannot be easily coagulated.

2) Lyophobic colloids:
If there is repulsion between dispersion medium and dispersed phase, the colloids are called lyophobic colloids. Lyophobic means solvent hating. This type of colloids cannot be prepared simply by mixing. They have to be prepared by special methods.

These sols are readily precipitated by the addition of small amounts of electrolytes or by heating. The precipitate does not give back the colloidal sol by simple addition of the dispersion medium to it. So these are known as irreversible sols. Lyophobic sols need stabilising agent for their preservation.

If water is the dispersion medium these are called hydrophilic and hydrophobic colloids.

Question 122.
Name a substance whose molecules consist of lyophilic as well as lyophobic parts. Give its use in our dally life.
Answer:
Micells contain both lyophilic as well as lyophobic parts eg. soap. Soap is sodium or potassium salt of higher fatty acid and may be represented as R COO^– Na⁺. Sodium stearate is CH₃ (CH₂)₁₆ COO^– Na⁺ is a major component of many soaps when dissolved in water it dissociates into CH₃ (CH₂)₁₆ COO^– and Na⁺ ions. The CH₃(CH₂)₁₆ COO^– consists of two parts, the long non polar hydrocarbon chain CH₃(CH₂)₁₆ is water repelling and it is hydrophobic end. The polar COO^– group is water attracting and is hydrophilic end.

In our daily life when we use soap, the hydrophobic part of soap ion penetrates into the oil droplet and the hydrophilic part project out of the oil droplet.

Since the polar group can interact with water, the oil droplet surrounded hydrocarbon part is pulled into water and is removed from the dirty surface. Thus soap emulsifies the oil and washes it out.

Question 123.
Describe Bredig’s arc method of preparation of colloids with a neat diagram.
Answer:
This process involves dispersion as well as condensation. In this method Electric arc is struck between electrodes of the metal immersed in the dispersion medium. The intense heat produced vapourises the metal which then condenses to form particles of colloidal size. Colloidal sols of metals such as gold, silver, platinum, etc. are prepared by this method.

Question 124.
Name any four examples of colloids by chemical methods with necessary chemical equations.
Answer:

1. AS₂S₃ sol can be prepared by double decomposition method.
   AS₂O₃ + 3H₂S AS₂S₃ (sol) + 3H₂O
2. Sulphur sol can be prepared by oxidation method.
   SO₂ + 2H₂S 3S_(sol) + 2H₂O
3. Gold sol can be prepared by reduction method.
   2AuCl₃ + 3HCH0 + 3H₂O 2Au (sol) + 3HCOOH + 6HCl
4. Ferric hydroxide sol can be prepared by hydrolysis method.
   FeCl₃ + 3H₂O Fe(OH₃) + 3 HCl

Question 125.
Describe the purification of colloidal solutions by the phenomenon of dialysis with a neat diagram.
Answer:

Dialysis is a process of removing a dissolved substance from a colloidal solution using a suitable membrane. The apparatus used for this purpose is called dialyser. A bag of suitable membrane containing the colloidal solution is suspended in a vessel containing a continous flow of water. The bag made with animal membrane, or parchment paper or cellophane sheet allow the molecules or ions to diffuse through it into the water and pure colloidal solution is left behind in the bag.

Question 126.
Explain the formation of micelles with a neat sketch.
Answer:
Certain substances behave as strong electrolytes at low concentrations but at higher concentrations exhibit colloidal behaviour due to the formation of aggregates. These aggregated particles are called micelles or associated colloids.

For example soap is a sodium or potassium salt of higher fatty acid and may be represented as RCOO^– Na⁺ eg. sodium stearate CH₃ (CH₂)]₆ COO^– Na⁺, It dissociates into RCOO^– and Na⁺ ions when dissolved in water. The RCOO^– ions consists of two parts. The long hydrocarbon chain R (also called non polar tail) which is hydrophobic (water repelling) and a polar group COO^– which is hydrophilic.

The RCOO^– ions are therefore present with their COO^– group into water and hydrocarbon chains (R) staying away from it and remain at the surface. At critical micelle concentration the COO^– Ions are pulled into the bulk of the solution and are aggregated to form spherical shape with their hydrocarbon chain pointing towards the centre of sphere with COO^– part remaining outward on the surface of the sphere. The aggregate thus formed is known as ionic micelle.

Question 127.
Action of soap is due to emulsification and micelle formation. Comment.
Answer:
The cleansing action of soap is due to the fact that soap molecules form a micelle around the oil droplet in such away that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet.

Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled into water and is removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats.

Question 128.
Explain the phenomenon of Brownian movement giving reasons for the occurrence of this phenomena.
Answer:
The colloidal particles in a colloid are in continuous zig-zag motion all over the field of view. This continous zig-zag motion of colloidal particles in a colloid is called as Brownian movement.

The Brownian movement is due to the bombardment of the molecules of dispersion medium on colloidal particles (dispered phase). The motion is independent of the nature of the colloid but depends on the size of the particles and viscosity of the solution.

Question 129.
Name any four positively charged sols.
Answer:
Positively charged sols.

1. Hydrated metallic oxide sols e.g: Al₂O₃ . X H₂O, CrO₃ . X H₂O and Fe₂O₃ . X H₂O
2. Basic dye stuffs eg. methylene blue sol
3. Haemoglobin (blood)
4. Oxides e.g : TiO₂ sol

Question 130.
Name any four negatively charged sols.
Answer:

1. Metal sols e.g.: Copper, silver, gold sols.
2. Metallic sulphide sols e.g: AS₂S₃, Sb₂S₃, Cds etc.
3. Acid dye stuff sols, e.g: eosin congo red sols.
4. Sols of starch, gum, gelatin, clay, charcoal etc.

Question 131.
Explain the terms Helmholtz electrical double layer and zeta potential. What are their significances in the colloidal solutions ?
Answer:
In a colloid the colloidal particles acquire positive or negative charge by selective adsorption of ions on the surface of the colloidal particle. This positive or negative charge on colloidal particle attracts ions of opposite charge from the medium forming a second layer as shown below.
Agl/I^– K⁺
AgI/Ag⁺ I^–
The combination of the two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer. The first layer of ions is firmly held and is known as fixed layer while the second layer is mobile and is known as diffused layer. Between these two layers of opposite charges there develops a potential difference called electrokinetic potential or zeta potential.

The presence of equal and similar charges on colloidal particles is responsible for the stability of collodial solution. The repulsion between the colloidal particles having same charge keep them away and preventing them to come closer and coagulate.

Question 132.
Explain with a neat sketch the phenomenon of electrophoresis.
Answer:

When electric potential is applied across two platinum electrodes dipping in a colloidal solution the colloidal particles move towards one or the other electrode. The movement of colloidal particles under the applied emf is called electrophoresis postively charged particles move towards the cathode while negatively charged particles move towards the anode.

Question 133.
Explain the following terms.
i) Electrophoresis
ii) Coagulation
iii) Tyndall effect
Answer:
i) Electrophoresis:
When electric potential is applied across two platinum electrodes dipping in a colloidal solution the colloidal particles move towards one or the other electrode. The movement of colloidal particles under the applied emf is called electrophoresis. Positively charged particles move towards the cathode while negatively charged particles move towards the anode.

ii) Coagulation:
The process of setting down of colloidal particles is called coagulation or precipitation or flocculation of the sol. When the charge on colloidal particles is neutralised, the colloidal particles come nearer to form aggregates and settles down under the force of gravity.

iii) Tyndall effect :
When light is passed through a colloidal solution the colloidal particles scatter the light in all directions in space. This scattering of light illuminates the path of beam in the colloidal dispersion. This is known as Tyndall effect.

Question 134.
Explain the phenomenon observed,
i) When a beam of light is passed through a colloidal sol.
ii) An electrolyte, NaCl is added to hydrated ferric oxide.
iii) An electric current is passed through a colloidal solution.
Answer:
i) When a beam of light is passed through a colloidal sol the colloidal particles scatter light in all directions in space. This scattering of light illuminates the path of beam in the colloidal dispersion. This is known as Tyndall effect.

ii) When NaCl is added, the colloidal particles are precipitated. This is because that colloidal particles interact with ions carrying opposite charge to that present on themselves. Then neutralisation of charges on colloidal particles takes place leading to their coagulation.

iii) When electric current is passed through a colloidal solution the colloidal particles move towards oppositely charged electrodes, get discharged and finally precipitated.

Question 135.
Describe Cottrell smoke precipitator with a neat diagram.
Answer:
Smoke is a colloidal solution of solid particles such as carbon, arsenic compounds, dust etc. in air. The smoke before coming out of chimney is passed through a precipitator containing plates having a charge opposite to that carried by smoke particles. The particles on coming in contact with these plates lose their charge and get precipitated. The particles thus settle down on the floor of the chamber. The precipitator is called Cottrell precipitator.

Question 36.
Among NaCl, Na₂SO₄ Na₃ PO₄ electrolytes which is more effective for coagulation of hydrated ferric oxide sol and why ?
Answer:
Hydrated ferric oxide sol carry positive charge. So it can be coagulated by adding an electrolyte that can neutralise the positive charge. Among NaCl, Na₂SO₄ and Na₃PO₄, Na₃PO₄ is more effective for coagulation because from Na₃PO₄ the PO₄^3- ion formed can neutralize effectively. This is according to Hardy Schulze law which states that greater the valence of coagulating ion added the greater is its power to cause coagulation.

Question 137.
Discuss how a lyophilic colloid protect a lyophobic colliod.
Answer:
Lyophilic sols are more stable than lyophobic sols. In lyophilic colloids the particles are extensively solvated and covered by a layer of the dispersion medium. When a lyophilic colloid is added to a lyophobic sol the lyophilic particles form a protective layer around lyophobic particles and thus protect the latter from the action by electrolytes. Lyophilic colloids used for this purpose are called protective colloids.

Question 138.
Discuss the use of colloids in
i) Purification of drinking water
ii) Tanning
iii) Medicines
Answer:
i) Purification of drinking water :
Water obtained from natural sources contain colloidal particles. Alum is added to such water to coagulate the colloidal impurities and make water fit for drinking purposes.

ii) Tanning :
Animal skins are colloidal in nature. When the skin containing positively charged particles is soaked in tannin that contain negatively charged colloidal particles, mutual coagulation takes place. This makes the skin hard and the process is called Tanning.

iii) Medicines :
Most of the medicines are colloidal in nature. Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated, e.g : Argyrol is a silver sol used as an eye lotion. Colloidal antimony is used in curing Kala azar.

Question 139.
Define Gold number.
Answer:
The protective power of protective colloids (lyophilic colloids) is measured by gold number. It is defined as “the mass in mill-grams which protects the coagulation of 10 ml of a gold sol on adding 1 ml of 10% NaCl solution”.

Question 140.
How do emulsifiers stabilize emulsion ? Name two emulsifiers.
Answer:
Emulsions are unstable. For stablisation of emulsion, a third component called emulsifying agent is usually added. The emulsifiying agent forms an interfacial film between suspended particles and the medium.

1. Proteins are emulsifying agent for o/w type emulsions.
2. Heavy metal salts of fatty acids, long chain alcohols, lamp black, etc. are emulsifying agents for w/o type emulsion.

Long Answer Questions (8 Marks)

Question 141.
Explain the terms absorption, adsorption and sorption. Describe the different types of absorption. [AP ’15]
Answer:
Absorption :
Uniform distribution of a substance through out the bulk of the solid is called absorption. For example, when a sponge or chalk piece is dipped in water, it distributes uniformly throughout the sponge or chalk piece due to absorption.

Adsorption:
The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is termed as adsorption. The substance which accumulates on the surface is called adsorbate and the material oh which the adsorption take place is called adsorbent.

Sorption :
If adsorption and absorption takes place simultaneously, it is known as sorption.

Adsorption is two types 1) Physical adsorption or physisorption and 2) Chemical adsorption or chemisorption.
Physisorption:

1. It is due to the adsorption of a gas on a solid surface due to weak van der Waal’s forces.
2. It is universal and lacks of specificity.
3. The gases which have high critical temperatures adsorb easily.
4. It is reversible and increases with increase in pressure and decrease in temperature.
5. The enthalpy of physisorption is low.

Chemisorption:

1. If the adsorption of a gas on a solid surface takes due to chemical bonds, it is called chemisorption.
2. It involves high energy of activation.
3. Physical adsorption at low temperature may convert into chemical adsorption at high temperature.
4. It is highly specific in nature and is irreversible in nature.
5. The enthalpy of chemisorption is high since it involves chemical bond formation.

Question 142.
Discuss the characteristics of physical adsorption.
Answer:
Characteristics of physical adsorption.
i) Lack of specificity:
Since Van der Waal’s forces are universal thefe is no specificity for physical adsorption. Any gas can be absorbed on any solid.

ii) Nature of adsorbate :
The amount of gas adsorbed by a solid depends on the nature of gas. Easily liquefiable gases which have high critical temperatures are readily adsorbed. This is because van der Waal’s forces are stronger near the critical temperature.

iii) Reversibilty :
Physical adsorption of a gas by a solid is generally reversible. With increase in pressure and decrease in te’mperatue adsorption increases. This is because adsorption decreases the volume of gas and adsorption is exothermic. With decrease in pressure and increase in temperature adsorption decreases. This is accroding to Le chatelier’s principle.

iv) Surface area of adsorbent:
More the surface area more is adsorption. Finely divided metals and porous substances having more surface area adsorbs more amount of gases.

v) Enthalpy of adsorption:
Since van der Waal’s forces are weak the enthalpy of physical adsorption is quite low (20 – 40 kJ mol^-1).

Question 143.
Discuss the characteristics of chemisorption.
Answer:
Characteristics of Chemisorption :
i) High specificity :
Chemisorption is highly specific. It will occur only if there is possibility for the formation of chemical bonds between adsorbate and adsorbent.

ii) Irreversibility:
The chemical adsorption is due to compound formation i.e., due to chemical reaction. It is slow at low temperatures but increases with increase in temperature because the chemical reaction between adsorbent and adsorbate requires some energy of activation.
The physical adsorption taking place at low temperature may convert into chemical adsorption at a high temperature usually high pressures are favourable for chemisorption.

iii) Surface area:
With increase in surface area chemical adsorption increases.

iv) Enthalpy of adsorption :
Since chemisorption involves chemical bond formation, the enthalpy of chemisorption is also high (80 – 240 kJ mol^-1).

Question 144.
Compare and contrast the phenomenon of physisorption and chemisorption.
Answer:

Question 145.
What is an adsorption isotherm ? Discuss the phenomenon of adsorption of gases on solids with the help of Freundlich adsorption isotherm.
Answer:
The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve called as adsorption isotherm.

Freundlich adsorption isotherm :
It gives the empirical relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. The relationship can be expressed as
\(\frac{\mathrm{x}}{\mathrm{m}}\) = k . p^1/n (n > 1) ……………….. (1)

where x is the mass of gas adsorbed
m is the mass of adsorbent
p is pressure
k and n are constants which depend on the nature of gas and the adsorbent at a particular temperature.

The curves in the graph indicate that at constant pressure, there is decrease in physical adsorption with increase in pressure.
Taking logarithms of equation (1)
log \(\frac{\mathrm{x}}{\mathrm{m}}\) = log k + \(\frac{\mathrm{1}}{\mathrm{n}}\) log p …………… (2)
on plotting log \(\frac{\mathrm{x}}{\mathrm{m}}\) on y – axis and log p on x axis if a straight line is obtained Freundlich isotherm is valid. The slope of the straight line gives the value of \(\frac{\mathrm{1}}{\mathrm{n}}\). The intercept on n the y – axis give the value of log k.

The factor \(\frac{\mathrm{1}}{\mathrm{n}}\) can have values 0 and 1. Thus equation 2 holds good over a limited range of pressures when \(\frac{\mathrm{1}}{\mathrm{n}}\) = 0, \(\frac{\mathrm{x}}{\mathrm{m}}\) = constant
the adsorption is independent of pressure,
when \(\frac{\mathrm{1}}{\mathrm{n}}\) = 1, \(\frac{\mathrm{x}}{\mathrm{m}}\) = k.p i.e, \(\frac{\mathrm{x}}{\mathrm{m}}\) ∝ p. The adsorption varies directly with pressure. Freundlich adsorption isotherm fails at high pressures.

Question 146.
Give a detailed account of applications of adsorption.
Answer:
The adsorption phenomenon have a number of applications.

1. Production of high vacuum: The traces of air remained in a vessel after evacuation by a vacuum pump can be removed by adsorbing on charcoal to give high vacuum.
2. Gas masks : Gas masks consisting of activated charcoal or mixture of adsorbents adsorb poisonous gases during breathing. These are mainly used by coal miners.
3. Control of humidity : The mositure in air can be decreased by adsorbing it on silica gel or alumina gel thus the humidity in a room can be controlled.
4. Removal of coloured substances: Animal charcoal removes colours of impure coloured solution by adsorbing impurities.
5. Heterogeneous catalysis : Reactant molecules adsorb on the solid catalyst and decrease to activation energy. Thus the reaction proceed fastly.
6. Separation of Inert gases : The inert gases can be separated by adsorbing them on coconut charcoal at different tempe-ratures. This process depends on the difference in the degree of adsorption of gases.
7. In curing diseases : Several drugs kill germs by adsorbing on germs.
8. Froth floatation process: Sulphide ores are concentrated by this method.

Question 147.
What is catalysis ? How is catalysis classified ? Give four examples for each type of catalysis.
Answer:
Substances which accelerate the rate of a slow chemical reaction but whose chemical nature and mass remain unchanged after the reaction are known as catalysts and the phenomenon is known as catalysis.

Catalysis can be broadly divided into two groups.
I) Homogeneous catalysis: When the reactants and the catalyst are in the same phase, i.e., either in liquid or in gas phase, the process is said to be homogeneous catalysis.
Examples:
1) Oxidation of SO₂ to SO₃ in the presence of oxides of nitrogen.

SO₂, O₂ and the catalyst NO are all gases and are in same phase.

2) Hydrolysis of methyl acetate catalysed by H⁺.
CH₃COOCH₃CO + H₂O (l) CH₃ COOH (aq) + CH₃OH (aq)
Both reactants and the catalyst are in same liquid phase.

3) Hydrolysis of sugar catalysed by H⁺.

4) In the oxidation of oxalic acid with aqueous KMnO₄ in acid medium Mn²⁺ ion formed during the reaction act as catalyst.
2KMnO₄ (aq) + 3H₂SO₄ (aq) + 3H₂C₂O₄ (aq) →K₂SO₄ (aq) + 2MnSO₄ (aq) + 8H₂O (l) + 10CO₂ (g)

II) Heterogeneous catalysis: The catalytic process in which the reactants and the catalyst are in different phases is known as heterogeneous catalysis.
Examples:
1) Oxidation of SO₂ to SO₃ in the presence of Pt.

2) Synthesis of ammonia by the reaction of N₂ gas and H₂ gas in the presence of solid iron powder catalyst in Haber’s process.

3) Oxidation of ammonia to nitric oxide in the platinum gauze.

4) Hydrogenation of vegetable oils in the presence of solid nickel as catalyst.

Question 148.
Discuss the mechanism of heterogeneous catalysis.
Answer:
As per old theory the reactant molecules in the gaseous state or in solutions are adsorbed on the surface of the solid catalyst. The increase in concentration of the reactants on the surface increase the rate of reaction. The heat liberated during adsorption is also utilised in increasing the rate of reaction.
According to modern theory.

1. Reactant molecules diffuse on to the surface of catalyst.
2. The reactant molecules adsorb on the surface of the catalyst.
3. Chemical reaction takes place between adsorbed reactant molecules through formation of an intermediate.
4. The product molecules desorb from the catalyst surface making the surface available for more reaction to occur.
5. The product molecules diffuse away from the surface of the catalyst for the adsorption of fresh reactant molecules.

Question 149.
What are enzymes ? Explain in detail the enzyme catalysis with necessary examples.
Answer:
Enzymes are complex nitrogeneous organic . compounds which are produced by living plants and animals. They are mainly protein molecules of high molecular mass and form ‘ colloidal solutions in water. These catalyse numerous reactions that occurs in plants and animals. So they are termed as bio-chemical catalysts and the phenomenon is known as biochemical catalysis.
Mechanism of enzyme catalysed reaction :

There are a number of cavities present on the surface of colloidal particles of enzymes. These cavities are of characteristic shape and possess active groups such as -NH₂, – COOH, -SH -OH etc. These are the active centres on the surface of enzyme particles. The molecules of the reactant which have complimentary shape fit into these cavities just like a key fits into a lock. The reactant molecule combine with active groups forming an activated complex which decomposes into the products. These reactions may be considered to proceed in two steps.

1. To form an activated complex
   E + S → ES#
2. Decomposition of activated complex to give the product
   ES# → E + P

Examples:
1) Inversion of cane sugar in the presence of invertase.

2) Conversion of glucose into ethyl alcohol in the presence of zymase.

Question 150.
What are colloidal solutions ? How they are classified ? Give examples.
Answer:
A colloidal solution is a heterogeneous system in which one substance is dispersed (dispersed phase) as large particles in another substance (dispersion medium).

Colloids are classified on the basis of the following criteria.
i) Physical states of the dispersed phase and the dispersion medium.

ii) Nature of the interaction between the dispersed phase and the dispersion medium.
Depending upon the nature of interaction between the dispersed phase and dispersion medium the colloidal sols are classified into two types.
a) Lyophilic colloids:
If there is attraction between the dispersion medium and dispersed phase, the colloids are called as lyophilic colloids. Lyophilic means solvent loving or solvent attracting. These can be prepared by direct mixing and can be separated from one another by evaporating. These sols are reversible, so called reversible sols. These are stable and cannot be easily coagulated.

b) Lyophobic colloids :
If there is repulsion between dispersion medium and dispersed phase, the colloids are called lyophobic colloids. Lyophobic means solvent hating. This type of colloids cannot be prepared simply by mixing. They have to be prepared by special methods. These sols are readily precipitated by the addition of small amounts of electrolytes or by heating. The precipitate does not give back the colloidal sol by simple addition of the dispersion medium to it. So these are known as irreversible sols. Lyophobic sols need stabilising agents for their preservation.

If water is the dispersion medium these are called hydrophilic and hydrophobic colloids.

iii) Type of particles of the dispersed phase: :
Depending upon the type of the particles of the dispersed phase colloids are classified as multimolecular, macro. molecular and associated colloids.
a) Multimolecular colloids contain aggregates of atoms or molecules of dispersed phase to from particles of colloidal size, eg: Sulphur sol.

b) Macromolecular colloids contain dispersed phase of macromolecules of colloidal range. e.g. : Starch, cellulose, proteins and enzymes etc.

c) Associated colloids contain aggregates of long non polar hydrocarbon chain along with polar groups such as COOHSO₃H. The hydrocarbon chain is hydrophylic end and polar group is hydrophobic end. These aggregated particles formed are called micelles and also known as associated colloid.

Question 151.
How are colloids classified on the basis of the nature of interaction between a dispersed phase and a dispersion medium ? Describe an important characteristic of each class. Which of the sols need stabilising agents for preservation ?
Answer:
Nature of the interaction between the dis-persed phase and the dispersion medium.
Depending upon the nature of inter-action between the dispersed phase and dispersion medium the colloidal sols are classified into two types.
a) Lyophilic colloids:
If there is attraction between the dispersion medium and dispersed phase, the colloids are called as lyophilic colloids. Lyophilic means solvent loving or solvent attracting. These can be prepared by direct mixing and can be separated from one another by evaporating. These sols are reversible, so called reversible sols. These are stable and cannot be easily coagulated.

b) Lyophobic colloids :
If there is repulsion between dispersion medium and dispersed phase, the colloids are called lyophobic colloids. Lyophobic means solvent hating. This type of colloids cannot be prepared simply by mixing. They have to be prepared by special methods.

These sols are readily precipitated by the addition of small amounts of electrolytes or by heating. The precipitate does not give back the colloidal sol by simple addition of the dispersion medium to it. So these are known as irreversible sols. Lyophobic sols need stabilising agents for their preservation.

If water is the dispersion medium these are called hydrophilic and hydrophobic colloids.

Question 152.
What are micelles ? Discuss the mechanism of micelle formation and cleaning action of soap.
Answer:
Micelles and Mechanisim of micelle for-mation : Certain substances behave as strong electrolytes at low concentrations but at higher concentrations exhibit colloidal behaviour due to the formation of aggregates. These aggregated particles are called micelles or associated colloids.

For example soap is a sodium or potassium salt of higher fatty acid and may be represented as RCOC^– Na⁺ eg. sodium stearate CH₃ (CH₂)₁₆ COO^– Na⁺. It dissociates into RCOO^– and Na⁺ ions when dissolved in water. The RCOO^– ions consists of two parts. The long hydrocarbon chain R (also called non polar tail) which is hydrophobic (water repelling) and a polar group COO^– which is hydrophilic.

The RCOO^– ions are therefore present with their COO^– group into water and hydrocarbon chains (R) staying away from it and remain at the surface. At critical micelle concentration the COO^– Ions are pulled into the bulk of the solution and are aggregated to form spherical shape with their hydrocarbon chain pointing towards the centre of sphere with COO^– part remaining outward on the surface of the sphere. The aggregate thus formed is known as ionic micelle.

Cleaning Action:
The cleansing action of soap is due to the fact that soap molecules form a micelle around the oil droplet in such away that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet.

Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled into water and is removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats.

Question 153.
Describe the properties of colloids with necessary diagrams wherever necessary.
Answer:
1) Colligative properties:
Since the colloidal particles are aggregates of small particles, the number of particles in colloidal solution is comparatively less than those in true solution. So the values of colligative properties such as lowering of vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure are lesser than those shown by true solution at same concentration.

2) Optical properties :
When a beam of light is passed through a colloidal solution the colloidal particles scatter the light in all directions. This scattering of light illuminates the path of beam in colloidal dispersion. This is known as tyndall effect. The bright cone of the light is called Tyndall cone.

Tyndall effect is observed only when the following two conditions are satisfied.

1. The diameter of the colloidal particle should not be much smaller than the wave length of the light used.
2. The refractive indices of the dispersed phase and the dispersion medium should differ greatly in magnitude.
   

3) Colour :
Colour of colloidal solution depends on the wavelength of light scattered by the colloidal particles. The wavelength depends on the size and shape of the colloidal particle. Large particles scatter shorter wavelength light and vice versa. Spherical particles scatter red light and disc like particles scatter blue light.
The colour of the colloid changes from the direction it is viewed.

4) Kinetic property:
The colloidal particles are in continous zig-zag motion due to bombardment of the molecules of the dispersion medium on collodial particles.

The motion of colloidal particle is independent of the nature of the colloid but depends on the size of the particles and viscosity of the solution.

5) Electrical properties:
Colloidal particles. always carry an electric charge. The charge on the colloidal particle is considered as a fixed layer. It attracts the opposite charge in the solution which forms a second diffused layer. Between these two oppositely charged layers a potential difference arises. This is known as electrokinetic or zeta potential.

When an emf is applied the colloidal particles move towards cathode or anode depending on the charge present on it. It is known as electrophoresis.
If the movement of colloidal particles is arrested by some suitable means, the dispersion medium moves in the opposite direction. This is known as electro-osmosis.

6) Coagulation or precipitation :
If the charge on the colloidal particles is neutralised by adding a suitable oppositely charged ion, the colloidal particles coagulates or precipitates.

Question 154.
What are emulsions ? How are they classified ? Describe the applications of emulsions. [AP ’17 ; TS ’16]
Answer:
Emulsions are liquid – liquid colloidal system. The dispersion of finely divided droplets in another liquid emulsion. The emulsions are two types.

1. Oil dispersed in water (o/w)
2. Water dispersed in oil (w/o)

a) In o/w type emulsions water is the dispersion medium and oil is dispersion phase eg. milk and vanishing cream.
b) In w/o type emulsions oil is the dispersion medium and water is dispersion phase, eg ; butter

Applications of emulsions:
Emulsions are useful
(a) in the digestion of fats in intestines
(b) in washing processes of clothes and crockery
(c) in the preparation of lotions, creams, ointments in pharmaceuticals and cosmotics
(d) in the extraction of metals
(e) in the conversion of cream into butter by churning.

Intext Questions – Answers

Question 1.
Write any two characteristics of chemisorption.
Answer:

1. Chemisorption is highly specific. It will occur only when there is possibility of chemical bonding between adsorbent and adsorbate.
2. Enthalpy of chemisorption is high (80 – 240 kJmol^-1) since it involves bond formation.

Question 2.
Why does physisorption decrease with increase in temperature ?
Answer:
Adsorption is exothermic process. So according to Le-Chatelier’s principle physical adsorption occurs readily at low temperature and decreases with increasing temperature.

Question 3.
Why are finely powdered substances more effective adsorbents than their non powdered crystal forms ?
Answer:
The extent of adsorption increases with the increase of surface area of the adsorbent. Thus finely divided metals and porous substances having large surface areas are good adsorbents.

Question 4.
Hydrogen used in Haber’s process is obtained by reacting methane with steam in presence of NiO as catalyst. The process is known as steam reforming. Why is it necessary to remove CO formed in steam reforming when ammonia is obtained by Haber’s process ?
Answer:
The CO present in hydrogen act as poison to the molybdenum acting as promoter for iron which is used as catalyst. So it is necessary to remove CO from H₂ obtained by steam reforming during the manufacture of ammonia by Haber’s process.

Question 5.
Why is ester hydrolysis slow in the beginning but is fast after sometime ?
Answer:
Ester hydrolysis is slow in the beginning and becomes faster after sometime. This is because of the formation of acetic acid during the reaction which act as catalyst.

Question 6.
What is the role of desorption in the process of adsorption catalysts ?
Answer:
Desorption of reaction products from the surface of catalyst and there by making the surface available again for adsorption of reactants. Thus more reactions takes place.

Question 7.
What modification can you suggest in the Hardy – Schulze law ?
Answer:
Hardy-Schulze law states that greater the valence of the coagulating ion added, greater is its power to cause coagulation. Instead of valence by using charge we can say greater the charge on the coagulating ion added more is its coagulating power.

Question 8.
Why is it essential to wash the precipitate in gravimetric chemical analysis with wash liquid before drying and weighing it quantitatively ?
Answer:
The ions present in the solution will be adsorbed on the solid precipitate in gravimetric chemical analysis. This gives erronious results. On washing the precipitate with wash liquid before drying and weighing it quantitatively the adsorbed ions will be removed.

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 5th Lesson General Principles of Metallurgy Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 5th Lesson General Principles of Metallurgy

Very Short Answer Questions (2 Marks)

Question 1.
What is the role of depressant in froth floatation ?
Answer:
It is possible to separate a mixture of two sulphide ores by adjusting proportion of oil to water or by using depressants in froth floatation process. For example, in the case of an ore containing ZnS and PbS, the depresent used is NaCN. It selectively prevents ZnS from coming to the froth but allows PbS to come with froth. NaCN forms a layer of Na₂ [Zn(CN)₄] on the surface of ZnS.

Question 2.
Between C and CO, which is a better reducing agent at 673K ?
Answer:
At 673K, CO is a better reducing agent than C.

Question 3.
Name the common elements present in the anode mud in the electrolytic refining of copper.
Answer:
Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum.
These elements are so present because they are less basic and less reactive.

Question 4.
State the role of silica in the metallurgy of copper. [IPE ’14]
Answer:
Silica removes FeO present as an impurity in the form of slag FeSiO₃.

Question 5.
Explain “poling”. [AP ’16, ’15]
Answer:
Poling is a metal refining process. The molten metal is stirred with logs (poles) of green wood. The impurities are removed as gases. Blister copper is purified by this method. The reducing gases evolved from the wood, prevent the oxidation of copper.

Question 6.
Describe a method for the refining of nickel.
Answer:
Nickel is refined by vapour phase refining.
Mond’s process: In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetra carbonyl.

The carbonyl is subjected to higher temperature so that it is decomposed giving the pure metal.

Question 7.
How is cast iron different from pig iron ?
Answer:
The iron obtained from blast furnace contains about 4% carbon and many impurities in smaller amount (eg : S, P, Si, Mn). This is known as pig iron.

Cast iron is different from pig iron and is made by melting pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content (about 3%) and is extremely hard and brittle.

Question 8.
What is the difference between a mineral and an ore ?
Answer:
Minerals are naturally occurring chemical compounds in the earth’s crust. Out of many minerals of the metal, only a few minerals are chemically and commercially viable, to be used as sources for the extraction of the metal. Such minerals are known as ores.

Question 9.
Why copper matte is put in silica lined converter?
Answer:
Iron is converted to slag by heating copper ore with silica. Iron silicate and copper are produced in the form of copper matte. This contains Cu₂S and FeS.
When O₂ is passed it converts the FeS into FeO.
2FeS + 3O₂ → 2FeO + 2SO₂
Silica is used to remove FeO as iron silicate slag.
FeO + SiO₂ → Fe SiO₃

Question 10.
What is the role of cryolite in the metallurgy of aluminium? [TS ’16, ’15]
Answer:
The role of cryolite is to

1. increase electrical conductivity,
2. to dissolve alumina,
3. to lower the melting point of alumina.

Question 11.
How is leaching carried out in the case of low grade copper ores ?
Answer:
Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria. The solution containing Cu⁺² is treated with scrap iron (or) H₂.
Cu⁺²_(aq) + H_2(g) → Cu_(s) + 2H⁺_(aq)

Question 12.
Why is zinc not extracted from zinc oxide through reduction using CO ?
Answer:
Zinc is not extracted from zinc oxide through reduction by using CO because ∆G value for the reduction reaction is + ve.
2ZnO + 2CO → 2Zn^– + 2CO₂, ∆G = + 200 kJ

Question 13.
Give the composition of the following alloys. [AP & TS Mar. 19; (Mar. 2018/1 7; IPE 14)]
a) Brass
b) Bronze
c) German silver
Answer:
a) Brass contains Cu 60% and Zn 40%.
b) Bronze contains Cu 60-80% and Sn 20-40%.
c) German silver contains Cu 25-40%, Zn 25-35%, Ni 40-50%.

Question 14.
Explain the terms gangue and slag.
Answer:
Ore is usually contaminated with earthy materials and undesired chemical compounds. These are collectively known as gangue or matrix.
Flux reacts with gangue forming slag. Slag can be removed in the liquid form as it has lower melting point than gangue.
eg: Flux + gangue → slag
FeO + SiO₂ → FeSiO₃

Question 15.
How is Ag or Au obtained by leaching from the respective ores ?
Answer:
In the metallurgy of silver and that of gold, the respective metals are leached with a dilute solution of NaCN or KCN in the presence of air. From the leached solution, the metal is obtained through displacement by zinc.

Question 16.
What are the limitations of Ellingham diagram ?
Answer:
Ellingham diagram normally consists of plots of ∆_fG° vs T for formation of oxides of elements. Such diagrams help us in predicting the feasibility of thermal reduction of an ore.
Limitations:

1. The graph simply indicates whether a reaction is possible or not. i.e., the tendency of reduction with a reducing agent is indicated. It does not say about the kinetics of the reduction process.
2. The interpretation of ∆G^⊖ is based on K. (∆G^⊖ = – RT In K). Thus it is presumed that the reactants and products are in equilibrium. This is not always true because the reactant / product may be solid.

Question 17.
Write any two ores with formulae of the following metals: [TS ’15; IPE ’14]
a) Aluminium
b) Zinc
c) Iron
d) Copper
Answer:

Question 18.
What is matte ? Give its composition. [AP ’15]
Answer:
During the extraction of copper from copper pyrites, the product of blast furnace consists of Cu₂S and a little of FeS. This product is known as matte. This contains Cu₂S and FeS.

Question 19.
What is blister copper? Why is it so called? [Mar. 2018 – TS]
Answer:
During the extraction of copper from copper pyrites, when the matte formed is subjected to Bessemerisation, a liquid solution of copper is formed.
2Cu₂O + Cu₂S → 6Cu + SO₂.
It has blistered appearance due to evolution of SO₂ and so it is called blister copper.

Question 20.
Explain magnetic separation of impurities from an ore.
Answer:

If either the ore or the gangue is capable of being attracted by a magnetic roller, magnetic separation is carried out. The powdered ore is carried on a conveyer belt which passes over a magnetic roller. The magnetic substance collects near the magnetic roller while non-magnetic material will collect away from the roller.
Ex : Haematite containing tin stone impurity is purified by this method.

Question 21.
What is flux ? Give an example.
Answer:
During metallurgical operations, a foreign substance is added to the ore to remove the gangue. This substances called flux. This can be removed with the help of perforated ladder.
Flux + gangue → slag
In the extraction of copper SiO₂ is used as a flux to remove FeO.

Question 22.
Give two uses each of the following metals:
a) Zinc
b) Copper
c) Iron
d) Aluminium
Answer:
Uses:
a) Zinc :

1. Zinc is used for galvanising iron.
2. Zinc is used as reducing agent in the manufacture of paints, dye-stuff’s etc.
3. Used in large quantities in batteries.

b) Copper:

1. Used in several alloys, eg : Brass (Cu and Zinc)
2. Making wires in electrical industry.

c) Iron:

1. Cast iron is used for casting stoves, railway sleepers, etc.
2. Wrought iron is used making wires, bolts, agricultural implements.

d) Aluminium :

1. Aluminium foils are used as wrapers for chocolates.
2. Aluminium is used in the extraction of chromium and manganese from their oxides.
3. Aluminium wires are used as electrical conductors.

Question 23.
Between C and CO, which is a better reducing agent for ZnO ?
Answer:
Coke is a better reducing agent for ZnO and not CO.

Question 24.
Give the uses of
a) Cast iron
b) Wrought iron
c) Nickel steel
d) Stainless steel
Answer:
a) Cast iron is used for casting stoves, railway sleepers, gutter pipes, toys etc.
It is used in the manufacture of wrought iron and steel.
b) Wrought iron is used in making anchors, wires, bolts, chains and agricultural implements.
c) Nickel steel is used for making cables, automobiles and aeroplane parts, pendulum and measuring tapes.
d) Stainless steel is used for cycles, auto-mobiles, utensils, pens etc.

Question 25.
How is aluminium useful in the extraction of chromium and manganese from their oxides ?
Answer:
Aluminium is a reducing agent because of its electropositive character, in the extraction of chromium and manganese from their oxides
Cr₂O₃ + 2Al → Al₂O₃ + 2Cr + heat
3Mn₃O₄ + 8 Al → 4Al₂O₃ + 9 Mn + heat

Short Answer Questions (4 Marks)

Question 26.
Copper can be extracted by hydrometallurgy but not zinc – explain.
Answer:
Zinc cannot be extracted by hydrometallurgy as it is a highly electropositive metal.

The E° value of Cu²⁺ / Cu is more than that of hydrogen while E° value of Zn⁺²/ Zn is less than that of H₂. Therefore, Cu⁺² can be reduced to Cu by H₂ but not zinc.
Cu⁺⁺; (aq) + H₂(g) → Cu(s) + 2H⁺ (aq)

Question 27.
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction ?
Answer:
Pyrites cannot be reduced by carbon (or) hydrogen because the ∆G⁰ of Cu₂S is greater than ∆G⁰ of Cu₂O. Hence the sulphide ore is first converted into oxide and then reduced.
Copper pyrites contains iron as impurity. The iron can be removed only by using silica with which it forms a slag.
FeO + SiO₂ → FeSiO₃

Question 28.
Explain Zone refining.
Answer:
Zone refining is based on the principle that the impurities are more soluble in the melt s than in the solid state of the metal.

A circular mobile heater is fixed at one end of a rod of the impure metal. The molten zone moves along with the heater which is moved forward. As the heater moves for-ward, the pure metal crystallises out of the rnetal and impurities pass on into the adjacent molten zone.

As this process is repeated, the impurities get concentrated at one end. This end is cut off. This method is very useful for producing semiconductor grade metals of very high purity, eg : germanium, silicon, boron.

Question 29.
Write down the chemical reactions taking place in the extraction of zinc from zinc blende.
Answer:
Zinc blende
Concentrated by froth floatation
↓
Concentrated ore
Roasting in air – The concentrated ore is roasted in the presence of excess air at about 1200k to form zinc oxide
↓

Reduction of ZnO with coke at in a fire clay retort to zinc.metal.
↓

Purification by fractional distillation or electrolysis using impure zinc as anode, white sheet of pure zinc as cathode and ZnSO₄ solution acidified with dil. H₂SO₄ as electrolyte
↓
Pure zinc metal

Question 30.
Write down the chemical reactions taking place in different zones in the blast furnace during the extraction of iron.
Answer:
Oxide ores of iron are mixed with limestone and coke and fed into a blast furnace from its top.

Hot air is blown from the bottom of the furnace and burnt to give temperature upto 2200 K in the lower portion itself. In upper part, the temperature is lower, and the iron oxides coming from top are reduced in steps to FeO.
At 500 – 800 K (lower temp, range upper part of blast furnace)
3 Fe₂O₃ + CO → 2Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Fe₂O₃ + CO → 2FeO + CO₂
At 900 – 1500 K (higher temp. range)
C + CO₂ → 2CO
FeO + CO →Fe + CO₂
CaCO₃ → CaO + CO₂
CaO + SiO₂ → CaSiO₃
Limestone is decomposed to CaO which removes silicate impurity of the ore as CaSiO₃ slag.

Question 31.
How is alumina separated from silica in the bauxite ore associated with silica ? Give equations. [TS ’15]
Answer:
Bauxite ore associated with silica impurity is called white bauxite. It is purified by serpeck’s process. In this process the bauxite is mixed with coke and heated in the presence of N₂ at 2075 K . Silica is reduced to silicon and escapes as vapour.
Al₂O₃ + 3C + N₂ → 2AlN + 3CO

The AlN is hydrolysed to form aluminium hydroxide.
AlN + 3H₂O → Al(OH)₃ + NH₃
The Al(OH)₃ ppt is filtered, washed, dried and ignited then pure alumina is formed.
2Al(OH)₃ → Al₂O₃ + 3H₂O

Question 32.
Giving examples to differentiate roasting and calcination. [AP & TS ’16 ; IPE ’14] [Mar. 19, ’18 – TS]
Answer:
i) Calcination involves heating of the ore in the absence of air just below its fusion temperature. Volatile matter escapes leaving behind the metal oxide. Calcination makes the ore porous and is made in reverberatory furnace.
Fe₂O₃ . xH₂O (S) Fe₂O₃(s) + xH₂O (g)
ZnCO₃ (s) ZnO (s) + CO₂ (g)
CaCO₃ . MgCO₃ (s) CaO(s) + MgO (s) + 2CO₂ (g)

ii) Roasting :
In roasting, the ore is heated in a regular supply of air in a furnace, at a temperature, below the melting point of the metal.
The ore loses sulphur as its oxide leaving behind oxide of the metal. Roasting is done in reverberatory or blast furnace.
2 ZnS + 3O₂ → 2ZnO + 2SO₂
2 PbS + 3O₂ → 2PbO + 2SO₂

Question 33.
The value of ∆G° for the formation of Cr₂O₃ is – 540 kJmol^-1 and that of Al₂O₃ is -827 kJ mol^-1. Is the reduction of Cr₂O₃ possible with Al ?
Answer:
The height of the line in an Ellingham diagram indicates instability of the oxide since the higher the line, more positive the ∆G, the less spontaneous the formation of the oxide.
Cr₂O₃ + 2Al → 2Cr + Al₂O₃
2Cr(s) + \(\frac{3}{2}\) O₂(g) → Cr₂O₃(s)
2Al(s) + \(\frac{3}{2}\) O₂ (g) → Al₂O₃ (s)
The line for oxidation of Cr is above that of Al. This means Cr₂O₃ is less stable than Al₂O₃ at all temperatures and Al will be able to reduce Cr₂O₂ to Cr. Thus, the lower the ∆G, line on the Ellingham diagram, the more stable is the compound. Hence reduction of Cr₂O₃ by Al is energetically favourable.

Question 34.
What is the role of graphite rod in the electrometallurgy of aluminium ?
Answer:
In the electrometallurgy of aluminium, a fused mixture of alumina, cryolite and fluorspur is electrolysed using graphite rod as anode and graphite lined iron as cathode.

During electrolysis, Al is liberated at cathode and CO, CO₂ are liberated at anode.

If some other metal is used as anode, instead of graphite then O₂ liberated will not only oxidise the metal of the electrode but could also convert some of the Al liberated at cathode back to Al₂O₃.

So the role of graphite rod in the electrometallurgy of aluminium is to prevent the liberation of O₂ at anode which may oxidise some of the liberated aluminium back to Al₂O₃.

Question 35.
Outline the principles of refining of metals by the following methods.
a) Zone refining
b) Electrolytic refining
c) Poling
d) Vapour phase refining
Answer:
a) Zone refining :
Impurities are more soluble in the melt than in the solid state of the metal. A circular mobile heater is fixed at one end of a rod of the impure metal. The molten zone moves along with the heater which is moved forward. As the heater moves forward the pure metal crystallises out of the metal and impurities pass on into the adjacent molten zone. The process is repeated several times and the heater is moved in the same direction from one end to the other end. At one end, impurities get concentrated. This end is cut off. Ex : Germanium.

b) Electrolytic refining :
Impure metal is used as anode. A strip of the same metal in pure form is used as cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal. The required metal gets deposited on the cathode in the pure form. The metal, constituting the impurity goes as the anode mud.

c) Poling :
In this method, the molten metal is stirred with logs of green wood. Then, the impurities are removed either as gases or they get oxidised and form Scum over the surface of the molten metal.
Ex : Blister copper is purified by this method. The reducing gases, evolved from the wood, prevent the oxidation of copper.

d) Vapour phase refining:
In this method, the metal is converted into its volatile compound and collected elsewhere. It is then decomposed to give pure metal.
Mond process for refining nickel: In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetra carbonyl.

The carbonyl is subjected to higher temperature so that it is decomposed giving the pure metal.

Question 36.
Predict the conditions under which Al might be expected to reduce MgO.
Answer:
Aluminium can reduce MgO to Mg only at temperature above 1350°C. Only at T > 1350°C, AG’f becomes more than ∆G°f of Al₂O₃ as per Ellingham diagram used as cathode. Electrolyte is the acidified solution of the salt of the same metal. The required metal gets deposited on the cathode in the pure form. The metal, constituting the impurity goes as the anode mud.
Anode : M → Mⁿ⁺ + ne^–
Cathode : Mⁿ⁺ + ne^– → M (Pure metal)
Copper is refined using electrolytic method. Anodes are of impure copper and pure copper. Copper strips are taken as cathode. Acidified copper sulphate solution is the electrolyte. Result of electrolysis is the transfer of copper in pure form from anode to the cathode.
Anode : Cu → Cu⁺⁺ + 2e^–
Cathode : Cu⁺⁺ + 2e → Cu

Question 37.
Explain the purification of sulphide ore by froth floatation method. [ Mar. ’18 AP] [AP Mar. 19; (AP ‘ 17, 15; TS ‘ 15)]
Answer:
In this process ore is powdered and made into a slurry with water. A rotating paddle is used to agitate the suspension in presence of an oil. Froth is formed which carries mineral particles. To this slurry froth collectors and stabilisers are added. Collectors (pine oil, xanthate) enhance non – wettability of the mineral particles. Froth stabilisers (e.g. cresols, aniline) stabilise the froth. The mineral particles become wet by oils while gangue particles become wet by water. The froth is light and is skimmed off. The ore particles are then obtained from the froth.

Question 38.
Explain the process of leaching of alumina from bauxite.
Answer:
Leaching often used if the ore alone but not the gangue is soluble in same suitable solvent.

Bauxite is usually contains SiO₂, iron oxides, and Titanium oxide as impurity. The powdered ore is digested with a concentrated solution of NaOH at 473 – 523 K and 35 – 36 bar pressure. Al₂O₃ is leached out as sodium aluminate leaving the other impurities behind.
Al₂O₃ (S) + 2NaOH + 3 H₂O (l) → 2Na[Al(OH)₄] (aq)
The aluminate is alkaline in nature and is neutralised by passing CO₂ gas and hydrated Al₂O₃ is precipitated.

At this stage freshly precipitated hydrated Al₂O₃ is added. Precipitation of Al₂O₃. xH₂O is obtained.
2Na[Al(OH)₄] (aq) + CO₂(g) → Al₂O₃ . x H₂O (s) + 2NaHCO₃ (aq)
The sodium silicate remains in solution and the insoluble hydrated alumina is filtered; dried and heated to give pure Al₂O₃.
Al₂O₃. x H₂O (s) Al₂O₃(S) + xH₂O(g)

Question 39.
What is Ellingham diagram ?
What information can be known from this in the reduction of oxides ?
Answer:
a) Ellingham diagram consists of plots of \(\Delta \mathrm{G}_{\mathrm{f}}^{\mathrm{o}}\) vs T for formation of oxides of elemens.
2 x M (s) + O₂ (s) → 2 M_xO
In this reaction, the gaseous amount is decreasing from left to right due to the consumption of gases leading to negative value of ∆S which changes the value of ∆G in the equation ∆G = ∆H – T∆S subsequently shifts towards higher side despite rising T.

b) Each plot is a staright line except when some change in phase (s → liq or liq → g) takes place. The temperature at which such change occurs is indicated by an increase in the slope on +ve side.

c) There is a point in a curve below which ∆G is negative. So M_xO is stable. Above this point M_xO decomposes on its own.

d) The height of the line in Ellingham diagram indicates instability of the oxide.

Uses : Ellingham diagram provides a basis for considering the choice of reducing agent in the reduction of oxides such diagram helps in predicting the feasibility of thermal reduction of an ore.

Question 40.
How is copper extracted from copper pyrites ?
Answer:
In the Ellingham diagram, the 4 Cu + O₂ → 2 Cu₂O line is at the top. So it is quite easy to reduce Cu₂O to Cu directly by heating with coke.
Most of the Cu ores are sulphides and some contain iron.
Ores :

1. Copper iron pyrites Cu FeS₂
2. Copper glance Cu₂S

First stage : Sulphide ores are purified by froth floatation.

Second stage :
Roasting of the sulphide ore.
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2FeS + 3O₂ → 2FeO + 2SO₂
In actual practice, the ore is heated in a reverberatory furnace after mixing with silica. FeO is removed by using silica as flux.
FeO + SiO₂ → FeSiO₃
Copper is produced in the form of copper matte. This contains Cu₂S and FeS.

Third stage :
Copper matte is then charged into silica lined convertor. Hot air is blown and some silica is added. FeS is converted to FeO. Cu₂S / Cu₂0 is converted to metallic copper.
2FeS + 3O₂ → 2FeO + 2SO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2Cu₂0 + Cu₂S → 6 Cu + SO₂
The solidified copper has blister appearance due to evolution of SO₂. Hence it is called blister copper.

Fourth stage :
Copper is refined using electrolytic method. Anodes are of impure copper and pure copper strips are taken as cathode. Electrolysis is carried out using acidified CuSO₄ solution.
Anode : Cu → Cu⁺⁺ + 2e^–
Cathode : Cu⁺⁺ + 2e → Cu
Flow diagram :
Copper pyrites CuFeS₂
Froth floatation
↓
Concentrated ore
Roasting (volatile impurities escape) Formation of Cu₂S and FeS.
↓
Roasted ore
Mixed with coke and sand
FeS eliminated as FeSiO₃
↓
Matte Cu₂S + FeS
Hot air and sand
Cu₂O and Cu₂S give copper
↓
Blister copper
Refining poling and electrolysis
↓
Pure copper metal.

Question 41.
Explain the extraction of zinc from zinc blende.
Answer:
Ores :

1. Zinc blende ZnS
2. Calamine ZnCO₃

Zinc blende is purified by froth floatation. Now the concentrated ore is subjected to roasting.
2ZnS + 3O₂ → 2ZnO + 2SO₂
Zinc oxide is reduced with coke.

The metal is distilled and collected by rapid chilling. Very pure zinc is obtained by electrolysis. Acidified CuSO₄ solution is taken as electrolyte. Impure zinc is taken as anode and pure zinc is taken as anode. On passing current pure zinc gets deposited on the cathode.
Extraction of Zinc :.
Zinc blende ZnS
Cone, by froth floatation
↓
Concentrated ore
Roasting in air
↓
ZnO
Reduction with coke at 1673 K
↓
Zinc
Purification by fractional distillation / electrolysis
↓
Pure zinc metal

Question 42.
Explain smelting process in the extraction of copper.
Answer:
The copper pyrites is heated in a reverberatory furnace after mixing with silion the furnace, iron oxide slags as iron silicate and copper is produced in the form of copper matte. This Contains Cu₂S and FeS.
FeO + SiO₂ → FeSiO₃ (Slag)
Copper matte is then charged into silica lined converter. Some silica is also added and hot air blast is blown to convert remaining FeS to FeO and Cu₂S / Cu₂O to the metallic copper.
2FeS + 3O₂ →2FeO + 2SO₂
2 Cu₂S + 3O₂ → 2 Cu₂O + 2SO₂
FeO + SiO₂ → FeSiO₃
2Cu₂0 + Cu₂S → 6 Cu + SO₂

Question 43.
ExplaIn electrometallurgy with an example.
Answer:
Electrometallurgy :
Metallurgy involving
the use of electric are furnace, electrolysis and other electrical operations is called electrometallurgy.

Electrolytic reduction of alumina :
Al₂O₃ is mixed with Na₃AlF₆ and fluorspar CaF₂. The function cryolite is to lower the melt ing point of Al₂O₃ and CaF₂ increases electrical conductivity. Electrolysis is carried out in an iron take lines inside with graphite which acts as cathode. A number of carbon rods, suspended in the electrolyte acts as anode. The temperature is maintained at 1175 to 1225 K and current is passed. Following reactions takes place.
Na₃AlF₆ → 3 NaF₃ + AlF₃
4 AlF₃ → 4 Al⁺³ + 12F^–
At cathode : 4Al⁺³ + 12e → 4 Al
At anode: 12F^– → 6F₂ + 12e
F₂ formed at anode reacts with alumina and form AlF₃.
2 Al₂O₃ + 6F₂ → 4 AlF₃ + 3O₂
Anode : Graphite
Cathode : Steel
At anode : C + O^2- → CO(g) + 2e^–
C (s) + 2O^2- → CO₂ (g) + 4e^–
At cathode : Al³⁺ (melt) + 3e^– → Al (l)
The oxygen liberated at anode reacts with the carbon of anode producing CO and CO₂.
Therefore, anode has to be replaced periodically.

The overall reaction is
2 Al₂O₃ + 3C → 4 Al + 3CO₂
This process is known as Hall – Heroult process.

Question 44.
Explain briefly the extraction of aluminium from bauxite.
Answer:
1) Bauxite is digested with hot cone. NaOH at 523K. Al₂O₃ dissolves. Impurities are filtered off.
Al₂O₃ + 2 NaOH + 3H₂O (l) → 2Na [Al(OH)₄] aq
The aluminate is alkaline in nature and is neutralised by passing CO₂ gas. Al₂O₃ is precipitated.
2Na [Al (OH)₄] (aq) + CO₂ (g) → Al₂O₃. x H₂O + 2NaHCO₃.
The sodium silicate remains in the solution and the insoluble hydrated alumina is filtered, dried and heated to give pure Al₂O₃.
Al₂O₃. X H₂O Al₂O₃ (s) + xH₂O(g)

2) The purified Al₂O₃ is mixed with Na₃ AlF₆ or CaF₂ which lower the melting point of the mix and also increases conductivity. The fused matrix is electrolysed.

Steel vessel with lining of carbon acts as cathode and graphite anode is used. The overall reaction may be 2Al₂O₃ + 3C → 4Al + 3CO₂.
Reactions: At cathode Al³⁺ (melt) + 3e^– → Al
At anode C(s) + O^2- (melt) → CO (g) + 2e^–
C(s) + 2O^2- (melt) → CO₂ (g) + 4e^–
Flow Diagram :
Bauxite Al₂O₃ . 2H₂O digested with hot cone. NaOH Al₂O₃ dissolves – impurities eliminated
Na [Al (OH)₄]
CO₂ bubbled
Al (OH)₃ PPt
heated at 1473K
Pure alumina
electrolysis of alumina mixed with molten cryolite or
CaF₂ at 1200 K using carbon lining of cell – cathode
carbon rods – anode.
Pure aluminium metal.

Long Answer Questions (8 Marks)

Question 45.
The choice of a reducing agent in a particular case depends on thermodynamic factor. Explain with two examples.
Answer:
Consider the reduction of Cr₂O₃ by Al.
Cr₂O₃ + 2 Al → 2 Cr + Al₂O₃
2Cr (s) + \(\frac{3}{2}\) O₂(g) → Cr₂O₃ (s)
and 2 Al (s) + \(\frac{3}{2}\) O₂ (s) → Al₂O₃(s)
The line for oxidation of Cr is above that of Al. This means Cr₂O₃ is less stable than Al₂O₃ at all temperatures and Al will be able to reduce Cr₂O₃ to Cr. Thus, the lower the ∆G line on the Ellingham diagram, the more stable the compound. Hence, reduction of Cr₂O₃ by Al is energetically favourable.

Ex : 2 – The lower the line in Ellingham diagram, the more stable the oxide. Hence if Mg is heated with ZnO, it will reduce to zinc, but the reverse redaction does not occur.
Mg + ZnO → MgO + Zn

Question 46.
Discuss the extraction of zinc from zinc blende.
Answer: R
Ores :

1. Zinc blende ZnS
2. Calamine ZnCO₃

Zinc blende is purified by froth floatation. Now the concentrated ore is subjected to roasting.
2ZnS + 3O₂ → 2ZnO + 2SO₂
Zinc oxide is reduced with coke.

The metal is distilled and collected by rapid chilling. Very pure zinc is obtained by electrolysis. Acidified CuSO₄ solution is taken as electrolyte. Impure zinc is taken as anode and pure zinc is taken as anode. On passing current pure zinc gets deposited on the cathode.
Extraction of Zinc :.
Zinc blende ZnS
Cone, by froth floatation
↓
Concentrated ore
Roasting in air
↓
ZnO
Reduction with coke at 1673 K
↓
Zinc
Purification by fractional distillation / electrolysis
↓
Pure zinc metal

Question 47.
Explain the reactions occurring in the blast furnance in the extraction of iron.
Answer:
In the Blast furnace, reduction of iron oxides takes place in different temperature ranges. Oxides ores of iron are mixed with limestone and coke and fed into blast furnace from its top. Hot air is blown from the bottom of the furnace and coke is burnt to give temperature upto about 2200 K in the lower portion itself. The CO and heat moves to upper part of the furnace.

At 500 – 800 K:-
3 Fe₂O₃ + CO → 2Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Fe₂O₃ + CO → 2FeO + CO₂

At 900- 1500K:-
C + CO₂ → 2CO
FeO + CO → Fe + CO₂
Limestone is also decomposed to CaO which removes silicate impurity as slag.
CaCO₃ → CaO + CO₂
CaO + SiO₂ → CaSiO₃ (slag)
The slag is in molten state and separates and from iron. This contains 4% C, and impurities (eg : P, S, Si, Mn). This is known as pig iron.

Flow Chart:
Haematite Fe₂O₃
Crushed ore is washed with stream of water.
↓
Concentrated ore
Roasting in air / calcination moisture and volatile impurities are w eliminated FeO changes to Fe₂O₃
↓
Calcinated ore
Calcincated ore, coke and limestone in 8 : 4 : 1 reduced by CO in blast 4, furnace SiO₂ remove as CaSiO₃
↓
Pigiron

Question 48.
Discuss the extraction of copper from copper pyrites.
Answer:
In the Ellingham diagram, the 4 Cu + O₂ → 2 Cu₂O line is at the top. So it is quite easy to reduce Cu₂O to Cu directly by heating with coke.
Most of the Cu ores are sulphides and some contain iron.
Ores :

1. Copper iron pyrites Cu FeS₂
2. Copper glance Cu₂S

First stage : Sulphide ores are purified by froth floatation.

Second stage :
Roasting of the sulphide ore.
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2FeS + 3O₂ → 2FeO + 2SO₂
In actual practice, the ore is heated in a reverberatory furnace after mixing with silica. FeO is removed by using silica as flux.
FeO + SiO₂ → FeSiO₃
Copper is produced in the form of copper matte. This contains Cu₂S and FeS.

Third stage :
Copper matte is then charged into silica lined convertor. Hot air is blown and some silica is added. FeS is converted to FeO. Cu₂S / Cu₂0 is converted to metallic copper.
2FeS + 3O₂ → 2FeO + 2SO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2Cu₂0 + Cu₂S → 6 Cu + SO₂
The solidified copper has blister appearance due to evolution of SO₂. Hence it is called blister copper.

Fourth stage :
Copper is refined using electrolytic method. Anodes are of impure copper and pure copper strips are taken as cathode. Electrolysis is carried out using acidified CuSO₄ solution.
Anode : Cu → Cu⁺⁺ + 2e^–
Cathode : Cu⁺⁺ + 2e → Cu
Flow diagram :
Copper pyrites CuFeS₂
Froth floatation
↓
Concentrated ore
Roasting (volatile impurities escape) Formation of Cu₂S and FeS.
↓
Roasted ore
Mixed with coke and sand
FeS eliminated as FeSiO₃
↓
Matte Cu₂S + FeS
Hot air and sand
Cu₂O and Cu₂S give copper
↓
Blister copper
Refining poling and electrolysis
↓
Pure copper metal.

Question 49.
Explain the various steps involved in the extraction of aluminium from bauxite.
Answer:
1) Bauxite is digested with hot cone. NaOH at 523K. Al₂O₃ dissolves. Impurities are filtered off.
Al₂O₃ + 2 NaOH + 3H₂O (l) → 2Na [Al(OH)₄] aq
The aluminate is alkaline in nature and is neutralised by passing CO₂ gas. Al₂O₃ is precipitated.
2Na [Al (OH)₄] (aq) + CO₂ (g) → Al₂O₃. x H₂O + 2NaHCO₃.
The sodium silicate remains in the solution and the insoluble hydrated alumina is filtered, dried and heated to give pure Al₂O₃.
Al₂O₃. X H₂O Al₂O₃ (s) + xH₂O(g)

2) The purified Al₂O₃ is mixed with Na₃ AlF₆ or CaF₂ which lower the melting point of the mix and also increases conductivity. The fused matrix is electrolysed.

Steel vessel with lining of carbon acts as cathode and graphite anode is used. The overall reaction may be 2Al₂O₃ + 3C → 4Al + 3CO₂.
Reactions: At cathode Al³⁺ (melt) + 3e^– → Al
At anode C(s) + O^2- (melt) → CO (g) + 2e^–
C(s) + 2O^2- (melt) → CO₂ (g) + 4e^–
Flow Diagram :
Bauxite Al₂O₃ . 2H₂O digested with hot cone. NaOH Al₂O₃ dissolves – impurities eliminated
Na [Al (OH)₄]
CO₂ bubbled
Al (OH)₃ PPt
heated at 1473K
Pure alumina
electrolysis of alumina mixed with molten cryolite or
CaF₂ at 1200 K using carbon lining of cell – cathode
carbon rods – anode.
Pure aluminium metal.

Intext Questions – Answers

Question 1.
Which ores can be separated by magnetic separation ?
Answer:
Ores in which one of the components is .agnetic can be concentrated eg : ores containing iron.
Ex : Haematite Fe₂O₃
Magnetite Fe₃O₄
Siderite FeCO₃
Iron pyrites FeS₂

Question 2.
What is the significance of leaching in the extraction of aluminium ?
Answer:
Leaching is significant as it helps in removing impurities like SiO₂, Fe₂O₃ etc. from bauxite ore.

Question 3.
Suggest a condition under winch magnesium could reduce alumina.
i) \(\frac{4}{3}\)Al + O₂ → 2/3 Al₂O₃
ii) 2 Mg + O₂ → 2MgO
Answer:
In the Ellingham diagram at the point of inter-section of Al₂O₃ and MgO, the ∆G^O becomes zero for the reaction.
2/3 Al₂O₃ + 2 Mg → 2 MgO + 4/3 Al.
Below that point, magnesium can reduce alumias.

Question 4.
The reaction Cr₂O₃ + 2Al → Al₂O₃ + 2Cr is thermodynamically feasible. Why does it not take place at room temperature.
Answer:
Certain amount of activation energy is essential even for such reactions which are thermodynamically feasible, therefore heating is required.

Question 5.
Is it true that under certain conditions, Mg can reduce Al₂O₃ and A1 can reduce MgO. What are those conditions ?
Answer:
Yes, below 1350°C Mg can reduce Al₂O₃ and above 1350°C, Al can reduce MgO. This can be inferred from ∆G^O vs T plots.

Question 6.
At a site, low grade copper ores are available and zinc and iron scraps are available. Which of the two scraps would be more suitable for reducing leached copper ore and why ?
Answer:
Zinc being above in electrochemical series, the reduction will be fast if zinc is used. But zinc is costlier than iron. So using iron scraps will be advisable and advantageous.

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material Lesson 3(a) Electro Chemistry Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material Lesson 3(a) Electro Chemistry

Very Short Answer Questions (2 Marks)

Question 1.
What is a galvanic cell or a voltaic cell ? Give one example.
Answer:
The cell that converts the chemical energy liberated during the redox reaction to electrical energy and exhibits an electrical potential is called galvanic cell or voltaic cell.
Ex : Daniel cell.

Question 2.
Write the chemical reaction used in the construction of the Daniel cell together with the half- cell reactions.
Answer:
In the Daniel cell the following redox reaction occurs.
Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu (s)

The reaction is a combination of the two half reactions whose addition gives the overall cell reaction.

1. Cu²⁺ (aq) + 2e^– → Cu (s) (reduction half reaction)
2. Zn (s) → Zn²⁺ (aq) + 2e^– (oxidation half reaction)

Question 3.
Name the two half-cell reactions that are taking place In the Daniel cell.
Answer:
The two half cell reactions that are taking place in the Daniel cell are as follows.

1. Cu²⁺(aq) + 2e^– → Cu (s) (reduction half reaction)
2. Zn (s) → Zn²⁺ (ag) + 2e^– (oxidation half reaction)

Question 4.
How is a galvanic cell represented on paper as per IUPAC convention ? Give one example.
Answer:
As per the IUPAC convention the anode is written on the left and the cathode on the right while representing the galvanic cell on paper. A galvanic cell is generally repre-sented by putting a vertical line in between the symbol of the metal (electrode) and formula of the electrolyte in the solution and putting a double vertical line in between the solutions (two electrolyte solutions). This indicates the connecting salt bridge (double vertical lines indicates salt bridge)
Examples: Zn (s) | Zn²⁺ (aq) || Cu²⁺ (aq) | Cu

Question 5.
Write the cell reaction taking place in the cell.
Cu(S) | Cu²⁺(aq) || Ag⁺ (aq) | Ag(S)
Answer:
As per the IUPAC convention the left hand side electrode is anode and right hand side electrode is cathode. At anode oxidation takes place while at cathode reduction takes place. These reactions can be shown as follows.

Question 6.
What is standard hydrogen electrode?
Answer:
Standard hydrogen electrode is used as reference electrode. Its potential was arbit-rarily fixed as zero. It consists of a platinum electrode coated with platinum black. The electrode is dipped in a solution of acid (us-ually 1M HCO and pure H₂ gas is bubbled through it at atmospheric pressure (or 1 bar). The concentration of H⁺ is unity (1M).

Question 7.
Give a neat sketch of standard hydrogen electrode
Answer:

Question 8.
What is Nernst equation? Write the equation for an electrode With electrode reaction
Mⁿ⁺ (aq) + ne^– \(\rightleftharpoons\) M(s).
Answer:
The value of electrode potential changes with the variation in the concentration of the ions. The equation which shows the relationship between the concentration of ions and electrode potentials is known as Nernst equation.
The Nernst equation for the electrode with electrode reaction

R = Gas constant (8.314 JK^-1 mol^-1)
F = Faraday constant (96487 C mol^-1)
T = Temperature in θ kelvin scale [Mⁿ⁺] is the concentration of the species Mⁿ⁺.

Question 9.
A negative E^θ indicates that the redox couple is… reducing couple than H⁺ / H₂, couple, (powerful or weak)
Answer:
Powerful.
More the negative reduction potential stronger the reducing agent.

Question 10.
A positive E^θ indicates that the redox couple is a weaker … couple than H⁺ / H₂ couple. (oxidising or reducing)
Answer:
Oxidising.

More positive E^θ indicates stronger oxidation property while more negative E^θ indicates stronger reducing property.

Question 11.
Write the Nernst equation for the EMF of the cell
Ni (s) / Ni²⁺ (aq) / / Ag⁺ (aq) / Ag
Answer:
The cell reaction is
Ni (s) + 2Ag⁺ (aq) → Ni²⁺ (aq) + 2Ag (s),
The Nernst equation is written as

R = Gas constant 8.3 14 JK mol^-1
F = Faraday constant 96487 C mol^-1
T = Temperature in kelvin scale

Question 12.
Write the cell reaction for which E_cell =

Answer:
Mg(s) + 2Ag⁺ (aq) → Mg²⁺ (aq) + 2Ag (s).

Question 13.
How is E^θ cell related mathematically to the equilibrium constant K_c of the cell reaction?
Answer:
\(\mathrm{E}_{\text {cell }}^\theta\) is related mathematically to equilibrium constant Kc as follows.
\(\mathrm{E}_{\text {cell }}^\theta\) = \(\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\) log K_C
R = Gas constant 8.3 14 JK mol^-1
F = Faraday constant 96487 C mol^-1
T = Temperature in kelvin scale
Kc = Equilibrium constant

Question 14.
How is Gibbs energy (G) related to the
Answer:
The emf of the cell E is related to Gibbs energy (G) as follows
Δ_rG = -nFE_cell
Δ_rG = Gibbs energy
n = number of electrons Involved in the reaction
F = Faraday constant 96487 C mol^-1
E_cell = cell emf

Question 15.
Define conductivity of a material. Give its SI units.
Answer:
Conductivity is the reciprocal of resistivity. It may also be defined as the conductance of one centimetre cube of the conductor. It is generally denoted by a Greek letter Kappa(K)
k = \(\frac{1}{\rho}\) = \(\frac{1}{\mathrm{R}}\left(\frac{l}{a}\right)\)
Units: The units of conductivity in SI units is S m^-1 or Ω^-1 m^-1.
Where S = Siemen

Question 16.
What is cell constant of a conductivity cell?
Answer:
In a conductivity cell the resistance of solution is given by the equation
R = \(\rho \frac{l}{A}\) = \(\frac{l}{\mathrm{KA}}\)
The quantity l/A is called cell constant denoted by the symbol G.

Question 17.
Define molar conductivity \(\Lambda_m\) and how is it related to conductivity (K)?
Answer:
Molar conductivity is defined as the conductance of a solution kept between the electrodes at unit distance apart and having area of cross – section large enough to accommodate sufficient volume of the solution that contains one mole of electrolyte. It is denoted by the symbol \(\Lambda_m\).
The molar conductivity \(\Lambda_m\) and conductivity K are related as
K = \(\frac{\Lambda_{\mathrm{m}}}{\mathrm{V}}\) (or) \(\Lambda_m\) = KV
V is the volume containing one mole of electrolyte.

Question 18.
Give the mathematical equation which gives the variation of molar conductivity \(\Lambda_m\) with the molarity (c) of the solution.
Answer:
The mathematical equation that gives the variation of molar conductivity \(\Lambda_m\) with the molaritý (c) of the solution is as follows. \(\Lambda_m\) (S cm² mol^-1)
= \(\frac{\mathrm{K}\left(\mathrm{S} \mathrm{cm}^{-1}\right)}{1000 \mathrm{~L} \mathrm{~m}^{-3} \times \text { molarity }\left(\mathrm{mol} \mathrm{L}^{-1}\right)}\)
If we use S cm^-1 as the units for K and mol cm^-3, the units of concentration, then the units for \(\Lambda_m\) are S cm² mol^-1. It can be calculated by using the equation
\(\Lambda_m\) (S cm² mol^-1)
= \(\frac{\mathrm{K}\left(\mathrm{S} \mathrm{cm}^{-1}\right) \times 1000\left(\mathrm{~cm}^3 / \mathrm{L}\right)}{\text { molarity }(\mathrm{mol} / \mathrm{L})}\)

Question 19.
State Kohlrausch’s law of independent migration of ions.
Answer:
The Kohlrausch’s law of independent migra-tion of ions states that “limiting molar con-ductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte”.

Question 20.
State Faraday’s first law of electrolysis. (AP Mar. 18, 16; TS 15; IPE 14)
Answer:
The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of current passing through the electrolyte (solution or melt). *
W ∝ Q

Question 21.
State Faraday’s second law of electrolysis. (Mar. 2018-TS, IPE 14)
Answer:
The amounts of different substances liberated, when the same quantity of current is passing through the electrolyte solution are proportional to their chemical equivalent weight (Atomic mass of metal ÷ number of electrons required to reduce the cation).
Mathematically: \(\frac{W_1}{W_2}=\frac{E_1}{E_2}\)

Question 22.
What are the products obtained at the platinum anode and the platinum cathode respectively in the electrolysis of fused
or molten NaCl?
Answer:
Electrolysis of fused or molten NaCl gives chlorine gas at anode and sodium metal at
cathode.
NaCl → Na⁺ + Cl^–
2C^– → Cl₂ (g) + 2e^– at anode
Na⁺ + e^– → Na (s) at cathode

Question 23.
Give the products obtained at the platinum electrodes (cathode and anode) when aqueous solution of K₂ SO₄ is electrolysed.
Answer:
In aqueous solution of K₂SO₄ the following reactions occur
K₂SO₄ → 2K⁺ + \(\mathrm{SO}_4^{2-}\)
The reactions taking place during electrolysis are
2H₂O → O₂ + 4H⁺ + 4e^– at anode.
4H₂O + 4e^– → 2H₂ + 4OH^– at cathode.
So at anode oxygen gas and at cathode hydrogen gas will be liberated,

Question 24.
Write the chemical equation corresponding to the oxidation of H₂O_(l) at the platinum anode.
Answer:
2H₂O → O₂ + 4H⁺ + 4e^– at Pt anode.

Question 25.
Give the chemical equation that represents the reduction of liquid water H₂O_(l) at the platinum cathode.
Answer:
4H₂O + 4e^– → 2H₂(g) + 4OH^– at Pt cathode.

Question 26.
What is a primary battery? Give one example. (AP 17)
Answer:
Primary batteries are the type óf batteries which become dead over a period of time and chemical reaction stops. They cannot be recharged or used again some common examples are dry cell, mercury cell etc.

Question 27.
Give one example for a secondary battery. Give the cell reaction.
Answer:
The most familiar example of secondary battery is lead storage battery.
Anode reaction

Question 28.
Give the cell reaction of nickel – cadmium secondary battery.
Answer:
The overall reaction of nickel – cadmium secondary battery is
Cd (s) + 2 Ni (OH)₃ (s) → CdO (s) + 2Ni (OH)₂ (s) + H₂O (l)

Question 29.
What is a fuel cell ? How is it different from a coventional galvanic cell ?
Answer:
Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells.
Galvanic cells directly convert chemical energy into electricity. In fuel cells reactants are fed continuously to the electrodes and products are removed contionuously from the electrolyte compartment.

Question 30.
Give the electrode reactions occurring at the anode and at the cathode in H₂, O₂ fuel cell.
Answer:
The electrode reactions occurring at the anode and at the cathode in H₂, O₂ fuel cell are as follows.
Cathode O₂ (g) + 2H₂O (l) + 4e^– → 4OH^– (aq)
Anode 2H₂ (g) + 4OH^– (aq) → 4H₂O (l) + 4e^–
The overall reaction is
2H₂ (g) + O₂ (g) → 2H₂O (l)

Question 31.
What is metallic corrosion ? Give one example.
Answer:
Corrosion may be defined as the process of slow conversion of metals into their unde-sirable compounds (usually oxides) by rea-ction with moisture and other gases present in the atmosphere.
Examples: The rusting of iron (iron oxide), tarnishing of silver (silver sulphide) deve-lopment of green coating on copper (copper carbonate).

Question 32.
Give the electro -chemical reaction that represents the corrosion or rusting of iron.
Answer:
Oxidation of iron takes place at one spot and that spot behaves as anode and the reaction is
Anode 2 Fe (s) → 2Fe²⁺ + 4e^–
Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen at it in presence of H⁺. This spot behaves as cathode.
Cathode O₂ (g) + 4H⁺ (aq) + 4e^– → 2H₂O (s)
The overall reaction is
2Fe (s) + O₂ (g) + 4H⁺ (aq) → 2Fe²⁺ (aq) + 2H₂O (l).

Short Answer Questions (4 Marks)

Question 33.
What are galvanic cells? Explain the work-ing of a galvanic cell with a neat sketch taking Daniell cell as example. (Mar. 2018 – AP & TS)( AP & TS ’15)
Answer:
The device in which chemical energy is converted into electrical energy is called galvanic cell or electrochemical cell or voltaic cell. In a galvanic cell, a redox reaction is carried in an indirect manner and the decrease in free energy during the chemical process is made to appear as electrical energy.

In the Daniell cell a zinc strip is dipped in the ZnSO₄ solution and a copper strip is dipped in the CuSO₄ solution taken in separate beakers. The two metallic strips which act as electrodes are connected by the conducting wires through Voltmeter. The two solutions are joined by an inverted U – tube known as salt bridge which contain an electrolyte such as KCl, KNO₃ or NH₄Cl along with gelatin or agar – agar to convert the electrolyte into semi solid paste.

The working of the cell can be understand by the following steps.

1. Zinc undergoes oxidation to form zinc ions.
   Zn (s) → Zn²⁺ (aq) + 2e^– oxidation
2. The electrons liberated during oxidation are pushed through the connecting wires to copper strip.
3. Copper ions move towards copper strip, pick up the electrons and get reduced to copper atoms which are deposited at the copper strip.
   Cu²⁺ (aq) + 2e^– → Cu (s) Reduction.

The transference of electrons from anode (oxidation) electrode i.e., zinc ele-ctrode) to cathode (reduction electrode i.e., copper electrode) leads to flow of electric current.

Question 34.
Give the construction and working of a standard hydrogen electrode with a neat diagram.
Answer:
The hydrogen electrode consists of platinum electrode coated with platinum black. The electrode is dipped in a solution of acid (us-ually 1M HCl) and pure H₂ gas is bubbled through it, at atmospheric pressure (or 1 bar). The concentration of H⁺ is unity (1M). For this electrode the potential is arbitrarily fixed as zero. This is known as standard hydrogen electrode. It can be represented as.
Pt (s) | H₂ (g) | H⁺ (aq)
Its potential value is zero volts at all temperatures.
The potential corresponds to the reaction.
H⁺ (aq) + e^– → \(\frac{1}{2}\)H₂(g)
This is called standard hydrogen electrode.

At 298 K the standard hydrogen electrode (SHE) is connected to another half cell and the emf of the cell (SHE II second half cell) gives the reduction potential of the other half cell. If the concentrations of the oxidised and the reduced forms of the species in the right hand half – cell are unity, then the cell emf is equal to standard electrode potential \(\mathrm{E}_{\mathrm{R}}^\theta\) of the given half cell.

Question 35.
State and explain Nernst equation with the help of a metallic electrode and a non metallic electrode.
Answer:
The equation that gives the quantitative relationship between the concentration of ions and electrode potentials is given by Nernst equation. For a general metallic electrode reaction.
Mⁿ⁺ (aq) + ne^– → M (s)
The Nernst equation can be written as

but the concentration of solid M is taken as unity

R is gas constant (8.314 JK^-1 mol^-1), F is Faraday constant (96487 C mol^-1), T is tem-perature in kelvin scale and [Mⁿ⁺] is the concentration of the species Mⁿ⁺. For a general non metallic electrode reaction.
X(g) + ne^– → X^n-(g).
The Nernst equation can be written as

But the concentration of gareous X is taken as unity

Question 36.
Explain with a suitable example the relation between the Gibbs energy of chemical reaction (G) and the functioning of the electrochemical cell.
Answer:
Electrical work done in one second in an electro chemical cell is equal to the electrical potential multiplied by the total charge passing. 1f we want to obtain maximum work from a galvanic cell then the charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs by a galvanic cell is equal to decrease in its Gibbs energy and therefore, if the emf of the cell is E and nF is the amount of charge passing and Δ_rG is the Gibbs energy of the reaction, then
Δ_rG = -nF E_cell
E_cell is an intensive parameter but Δ_rG is an extensive thermodynamic propery and the value depends on n. Eg: For the reaction
Zn (s) + Cu²⁺ (aq) () Zn²⁺ (aq) + Cu (s)
Δ_rG = -2F E_cell
If the concentration of each of the reacting species is unity then E_cell = -nF \(\mathrm{E}^\theta \text { cell }\) and we have
Δ_rG ^θ = -nF \(\mathrm{E}_{\mathrm{cel}}^\theta\)

Question 37.
On what factors the electrical conductance of an aqueous solution of electrolyte depends ?
Answer:
The conductivity of electrolytic (ionic) solutions depend on

1. the nature of the electrolyte.
2. size and solvation of the ions formed in the dissociation of the electrolyte.
3. the nature and viscocity of the solvent.
4. concentration of the electrolyte.
5. temperature (conductivity increases with increase in temperature).

Question 38.
How is molar conductivity of an aqueous electrolyte solution measured experimen-tally?
Answer:
The conductance of the solution is reciprocal of Its resistance. Therefore, if resistance of the solution is known, its conductance can be easily calculated. The resistance of the electrolytic solution is measured with the help of wheatstone bridge method. Its arrangement is shown in the following fig.

The setup consists of two resistances R₃ and R₄, a variable resistance R₁ and the conductivity cell having solution of unknown resistance R₂. An oscillator O (a source of a.c. power in the audio frequency range 550 to 5000 cycles per second) is connected to the bridge. P is a detector of null point (a head phone or an electronic device) and the bridge is balanced when no current passes through the detector. Under these conditions.
Unknown resistance R₂ = \(\frac{\mathrm{R}_1 \mathrm{R}_4}{\mathrm{R}_3}\)
Once the cell constant and the resistance of the solution in the cell are determined the conductivity of the solution is given by the equation

Molar conductivity \(\Lambda_{\mathrm{m}}\) = \(\frac{\mathrm{K}}{\mathrm{C}}\)

Question 39.
Explain the variation of molar conductivity with the change In the concentration of the electrolyte. Give reasons.
Answer:

1. Molar conductivity (\(\Lambda_{\mathrm{m}}\)) values of strong electrolytes is larger than those of weak electrolytes for the same concentration.
2. Molar conductivity of electrolytes, generally increase with dilution.
3. Relative increase in the value of \(\Lambda_{\mathrm{m}}\) for strong electrolyte is quite small as com-pared to that for weak electrolytes.

The degree of ionisatloñ of weak electrolytes is less in aqueous solutions. So their \(\Lambda_{\mathrm{m}}\) values are less. As the dilution increases the degree of ionisation of weak electrolyte also increases causing more and more ionisation. As a result the value of Am also increases significantlý.

Strong electrolytes ionise almost completely in aqueous solutions at all concentrations. Hence the values of their \(\Lambda_{\mathrm{m}}\) are generally high even at high concentrations. However in concentrated solutions of strong electrolytes there will be significant inter – ionic interactions which reduce the velocity of ions causing lower \(\Lambda_{\mathrm{m}}\) values. On increasing dilution, ions move away and inter ionic attractions decrease resulting in the increase iñ \(\Lambda_{\mathrm{m}}\) values.

Question 40.
State and explain Kohlrausch’s law of independent inigration of ions.
Answer:
Kohlrausch’s law of independent migration of ion states that at infine dilution when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductivity of electrolyte, irrespective of the nature of the other ion with which it is associated.

Thus the molar conductivity of an ele-ctrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions. If \(\lambda_{+}^{\circ}\) and \(\lambda_{-}^0\) represent the limiting molar conductivities of cation and anion respectively, Then the limiting molar conductivity of electrolyte at infinite dilution \(\Lambda_{\mathrm{m}}^{\circ}\), is given by

Where v₊ and v_– represent the number of positive and negative ions furnished by each formula unit of the electrolyte.

For example.

1. once formula unit of NaCl furnishes one Na⁺ and one Cl^– ion, therefore
   
2. one formula unit of BaCl₂ furnishes one Ba²⁺ and two Cl^– ions, therefore.
   

Question 41.
What is electrolysis? Give Faraday’s first law of electrolysis. (AP 15))
Answer:
The process of chemical decomposition of the electrolyte by the passage of electricity through its molten or dissolved state is called electrolysis.

Faraday’s first law of Electrolysis : This law states that ‘The mass of a substance liberated at the electrode is directly pro-portional to the quantity of electricity passed through the electrolyte.
m ∝ Q(or) m ∝ I × t (or) m = Z × I × t
Here Q = quantity of electricity
I = current in amperes
t = time in seconds
Z = constant of proportionality called electrochemical equivalent
If I = 1 ampere and t = 1 second then m = Z.

Thus the electrochemical equivalent of a substance is the amount of substance liber-ated at the electrode when current of one ampere is passed through the electrolyte for one second.

Question 42.
What are the products obtained at the cathode and the anode during the electrolysis of the following when platinum electrodes are used in the electrolysis.
a) Molten KCl
b) Aq. CuSO₄ solution
c) Aq. K₂SO₄ solution
Answer:

Question 43.
What are primary and secondary batteries? Give one example for each. (AP Mar. 19)
Answer:
Primary batteries are those which become dead over a period and the chemical reaction stops. They cannot be recharged or used again. Ex: Dry cell which is a compact form of Leclanche cell.

The cell consists of zinc container and acts as the anode. The cathode is a carbon (graphite) rod surrounded by powdered manganese dioxide and carbon. The space between the electrodes is filled by a moist paste of ammonium chloride and zinc chloride. The electrode reactions are as follows.
Anode Zn (s) → Zn²⁺ + 2e^–
Cathode MnO₂ + \(\mathrm{NH}_4{ }^{+}\) + e^– → MnO(OH) + NH₃

A secondary cell after its use can be recharged an can be used again.
Example : Lead storage battery. It consists of lead anode and a grid of lead packed with lead dioxide (PbO₂) as cathode. A 38% solution of sulphuric acid is used as electrolyte. The cell reactions are

The overall reaction is
Pb (s) + PbO₂(s) + 2H₂SO₄ (aq) → PbSO₄ (s) + 2H₂O
These reactions occur during discharge of the battery on charging the above reaction is reversed.

Question 44.
What are fuel cells ? How are they different from galvanic cells ? Give the construction of H₂, O₂ fuel cell.
Answer:
The cells which convert chemical energy of a fuel directly into electrical energy are called fuel cells. These are the Voltaic cells in which the fuels such as H₂, CO, CH₄, C₈H₈ etc. are used to generate electrical energy.

Galvanic cells convert the chemical energy liberated during the redox reaction to electrical energy. The emf of the galvanic cell is more if the intensity of the redox reaction is more. As the time proceeds the intensity of the redox reaction goes on decreasing and the cell becomes dead over a period of time when the cell reaction is completed. In fuel cells the reactants are fed continuously to the electrodes and products are removed continuously from the electrolyte compartment so that electrical energy is produced continuously.

In hydrogen fuel cell hydrogen and oxygen are bubbled through porous carbon electrodes placed in concentrated aqueous sodium hydroxide solution. Catalysts like finely divided platinum or palladium is incorporated into the electrodes for increasing the rate of electrode reactions. The electrode reactions are as follows.
Cathode : O₂ (g) + 2H₂O (l) + 4e,sup>- → 4OH^– (aq)
Anode: 2H₂ (g) + 4OH^– (aq) → 4H₂O (l) + 4e^–
The overall reaction
2H₂ (g) + O₂ (g) → 2H₂O (l)

Question 45.
What is metallic corrosion ? Explain it with respect to iron corrosion.
Answer:
Corrosion is a process of slow conversion of metals into oxides or other salts of the metal, generally the same as the compounds of their compounds by reaction with moisture and other gases present in the atmosphere.

Oxidation : Fe (s) → Fe²⁺ (aq) + 2e^–
Reduction: O₂ (g) + 4H⁺ (aq) + 4e^– → 2H₂O (1)
Atomospheric oxidation: 2Fe²⁺ (aq) + 2H₂O (1) + 1/2O₂ (g) → Fe₂O₃ (s) + 4H⁺ (aq)

Corrosion of iron is known as rusting. It occurs in presence of water and air. The corrosion of iron may be considered essentially as an electro chemical phenomenon. At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode. The reaction is

Long Answer Questions (8 Marks)

Question 46.
What are electro chemical cells ? How are they constructed ? Explain the working of the different types of galvanic cells.
Answer:
The device in which chemical energy is converted into electrical energy is called galvanic cell or electrochemical cell or voltaic cell.

In a galvanic cell, a redox reaction is carried out in an indirect manner and the decrease in free energy during the chemical process is made to appear as electrical energy. The indirect redox reaction is such that reduction and oxidation processes are carried out in separate vessels. The working of different types of galvanic cells can be understand by considering the Zn – CuSO₄ reaction as the basis of the cell reaction.

A zinc strip is dipped in zinc sulphate solution and a copper strip is dipped in copper sulphate solution taken in separate beakers. The two metallic strips which act as electrodes are connected by the conducting wire through a voltmeter. The two solutions are joined by a U – tube known as salt bridge which contain some electrolyte such as KCl, KNO₃ or NH₄Cl along with gelatin or agar – agar to convert it into semi – solid paste.

1. Zinc undergoes oxidation to form zinc ions.
   Zn (s) > Zn²⁺ (aq) + 2e^– (oxidation)
2. The electrons liberated during oxidation are passed through the connecting wire to copper strip
3. Copper ions move towards copper strip, pick up electrons and get reduced to copper atoms which are deposited on the copper strip.

At the zinc strip oxidation of zinc atoms takes place and becomes a source of ele-ctrons acquiring negative charge. It acts as anode, since oxidation occurs at it. At the copper strip reduction of copper ions takes place and acquires positive charge. It acts as cathode, since reduction takes place.

The flow of electrons from zinc strip to copper strip produce electric current through the outer circuit from copper to zinc strip which is indicated by the deflection in voltmeter.

Question 47.
What is electrical conductance of a solution? How is it measured experimentally ?
Answer:
The conduction through the electrolytes is due to the movement of ions produced by the electrolytes in their aqueous solution. It is generally called as electrolytic or ionic conductance.

Conductivity (K) or specific conductance is the reciprocal of resistivity. So if the resistance of the solution is known, its conductance can be easily calculated. The resistance of the electrolytic solution is measured with the help of Wheatstone bridge method. Its arrangement is shown in the figure.

The setup consists of two resistances R₃ and R₄, a variable resistance R₁ and the conductivity cell having solution of unknown resistance R₂. An oscillator 0(a source of a.c power in the audio frequency range 550 to 5000 cycles per second) is connected to the bridge. P is a detector of null point (a head phone or an electronic device) and the bridge is balanced when no current device) and the bridge is balanced when no current passed through the detector. Under these conditions
Unknown resistance R₂ = \(\frac{\mathrm{R}_1 \mathrm{R}_4}{\mathrm{R}_3}\)

Once the cell constant and the resistance of the solution in the cell are determined the conductivity of the solution is given by the equation.

Molar Conductivity \(\Lambda_{\mathrm{m}}\) = \(\frac{\mathrm{K}}{\mathrm{C}}\)

Question 48.
Give the applications of Kohlrausch’s law of independent migration of ions.
Answer:
1) Calculation of limiting molar couductivities of weak electrolytes: Kohlrauschs law is useful in determining the limiting molar conductivities of weak electrolytes. For example, the value of \(\Lambda_{\mathrm{m}}^{\circ}\) for acetic acid (CH₃COOH) can be calculated from the limiting molar conductivities of strong electrolytes like CH₃COONa, HCl and NaCl.

2) Calculation of degree of dissociation of weak electrolytes: Molar conductivity of weak electrolytes depends upon its degree of dissociation. With increase in dilution the degree of dissociation increases, so molar conductivity also increases. At infinite dilution the dissociation is complete and the molar conductivity of electrolyte becomes maximum attaining limiting molar conductivity (\(\Lambda_{\mathrm{m}}^{\circ}\)).
The degree of ionisation α is given by
α = \(\frac{\Lambda_{\mathrm{m}}^{\mathrm{c}}}{\Lambda_{\mathrm{m}}^{\circ}}\)
\(\Lambda_{\mathrm{m}}^{\mathrm{c}}\), is the molar conductivity of solution at any concentration C and \(\Lambda_{\mathrm{m}}^{\circ}\) is the limiting molar conductivity.

3) Calculation of dissociation constant of weak electrolyte: Dissociation constant of weak electrolyte can be calculated from its degree of dissociation at a given concentration.

4) Determining the solubility of sparingly soluble salts : The aqueous solutions of sparingly soluble salts are infinitely dilute solutions due to their extremely low solubility. At the same time they are also saturated solutions so the solubility can be calculated by the measurement of conductivity (K) and molar conductivity (\(\Lambda_{\mathrm{m}}\)) of the aqueous solution.

Question 49.
Give the different types of batteries and explain the construction and working of each type of battery.
Answer:
Batteries are two types:

1) primary batteries and
2) secondary batteries.

1) Primary batteries are those which become dead over a period and the chemical reaction stops. They cannot be recharged or used again.
Ex: Dry cell which is a compact form of Leclanche cell.

The cell consists of zinc container and acts as the anode. The cathode is a carbon (graphite) rod surrounded by powdered manganese dioxide and carbon. The space between the electrodes is filled by a moist paste of ammonium chloride and zinc chloride. The electrode reactions are as follows.
Anode Zn (s) → Zn²⁺ + 2e^–
Cathode MnO₂ + \(\mathrm{NH}_4^{+}\) + e^– → MnO(OH) + NH₃

2) Secondary cell : After use it can be recharged and can be used again.
Example : Lead storage battery. It consists of lead anode and a grid of lead packed with lead dioxide (PbO₂) as cathode. A 38%S solution of sulphuric acid is used as electrolyte. The cell reactions are

The overall reaction is
Pb (s) + PbO₂(s) + 2H₂ SO₄ (aq) → PbSO₄ (s) + 2H₂O (l)
These reactions occur during discharge of the battery. On charging the above reaction is reversed.

Questions Based On Numerical Data And Concept

Question 50.
The standard potentials of some electro-des are as follows. Arrange the metals in an increasing order of their reducing power.
1) K⁺ / K = -2.93 V
2) Ag⁺ / Ag = 0.80 V
3) Cu²⁺ / Cu = 0.34 V
4) Mg²⁺ / Mg = -2.37V
5) Cr³⁺ / Cr = -0.74 V
6) Fe²⁺ / Fe = -0.44 V
Answer:
Higher the value of oxidation potential or lower the reduction potential greater is the tendency of elements to oxidise and higher will be its reducing power. Thus the correct arrangement in decreasing order or their \(\mathrm{E}_{\text {red }}^\theta\)value is
Ag < Cu < Fe < Cr < Mg < K.

Question 51.
Calculate the emf of the cell at 25°C Cr | Cr³⁺ (0.1 M) || Fe²⁺ (0.01 M) || Fe²⁺(0.01M) / Fe given that \(\mathbf{E}_{\mathrm{Cr}^{3+} \mid \mathrm{Cr}}^0\) = -0.74V and \(\mathbf{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^0\) = -0.44 V
Answer:

Question 52.
Calculate the potential of a Zn – Zn²⁺ electrode in which the molarity of Zn²⁺ is 0.001 M. Given that \(\mathbf{E}_{\mathbf{Z n}^{2+} \mid \mathbf{Z n}}^{\ominus}\) = -0.76 V R = 8.314 JK^-1 mol^-1 ; F = 96500 C mol^-1
Answer:
Nernst equation

Question 53.
Determine \(\Delta G^{\ominus}\) for the button cell used in the watches. The cell reaction is

Answer:
In this cell zinc act as anode and silver electrode act as cathode

= +0.80 – (-0.76) = + 1.56 V
\(\Delta_r G^\theta\) = \(-\mathrm{nFE}_{\text {cell }}^\theta\) = -2 × 96500 × 1.56
= -3.01 × 10⁵ Cv (or) -301 kJ mol^-1

Question 54.
Calculate the emf of the cell consisting the following half cells

Answer:
The cell is
Al / Al³⁺ (0.001 M) || Ni²⁺ (0.50 M) / Ni
The cell reaction is

Question 55.
Determine the values of Kc for the follow-ing reaction

Answer:

Question 56.
Calculate the potential of the half – cell containing 0.1 M K₂Cr₂O₇, 0.2 M
\(\mathrm{Cr}^{3+}{ }_{(\mathrm{aq})}\) and 1 × 10^-4 M \(\mathrm{H}^{+}(\mathrm{aq})\)
The half – reaction

Answer:
The given equation is

Question 57.
Calculate K for the reaction at 298 K

Answer:
Calculation of \(\mathrm{E}_{\text {cell }}^\theta\)

Here zinc electrode is anode and copper electrode is cathode.
∴ \(\mathrm{E}_{\text {cell }}^\theta\) = + 0.34 – (-0.76) = 1.1 V

Calculation of K_c

Question 58.
Calculate the em! of the cell at 298 K
Sn_(s) | Sn²⁺ (0.05 M) || H⁺_(aq) (0.02 M) | H₂ 1 atm pt.
Given that \(\mathbf{E}_{\mathrm{Sn}^{2+} \mid \mathbf{S n}}^\theta\) = -0.144 V
Answer:
Cell reaction is
Sn_(s) + SH⁺ (aq) → Sn²⁺ (aq) + H₂ (g) n = 2
According to Nernst equation

Question 59.
Calculate the concentration of silver ions in the cell constructed by using 0.1 M concentration of Cu²⁺ and Ag⁺ ions. Cu and Ag metals are used as electrodes. The cell potential is 0.422 V

Answer:
The cell is Cu | Cu²⁺ Ag⁺ / Ag
The net cell reaction is

Question 60.
Calculate the em! of the cell with the cell reaction
Ni (s) + 2Ag⁺ (0.002 M) ) → Ni²⁺ (0.160 M) + 2Ag_(s) ; \(\mathbf{E}_{\text {cell }}^{\ominus}\) = 1.05 V.
Answer:
The cell reaction is
Ni (s) + 2Ag⁺ (aq) → Ni²⁺ (aq) + 2Ag (s)
Here Ni / Ni²⁺ electrode is anode and Ag⁺/ Ag electrode is cathode
Applying Nernst equation to above system we have

Question 61.
Cu²⁺ + 2e^– \(\rightleftharpoons\) Cu ; \(\mathbf{E}^{\ominus}\) = +0.34V
Ag⁺ + e^– \(\rightleftharpoons\) Ag ; \(\mathbf{E}^{\ominus}\) = + 0.80 V
For what concentration of Ag⁺ ions will the emf of the cell be zero at 25°C. The concentration of Cu²⁺ is 0.1 M (log 3.919 = 0.593).
Answer:
Since \(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\ominus}\) is larger than \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\ominus}\), reduction will occur at silver electrode and the cell is
Cu²⁺ / Cu || Ag⁺ / Ag
The cell reaction is
Cu(s) + 2Ag⁺(aq) → Cu²⁺ (aq) + 2 Ag (s)
According to Nernst equation

[Ag⁺] = (3.89 × 10^-16 × 0.1)<sup.1/2
= 6.222 × 10^-8.

Question 62.
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate molar conductance.
Answer:
Molar conductivity is given by = \(\frac{1000 \times K}{M}\)
= \(\frac{1000 \times 0.0248 \mathrm{~S} \mathrm{~cm}^{-1}}{0.20 \mathrm{M}}\)
= 12.4 cm² mol^-1

Question 63.
Calculate the degree of dissociation (α) of CH₃COOH at 298 K
Given that \(\Lambda_{\mathrm{CH}_3 \mathrm{COOH}}^{\infty}\) = 11.75 cm² mol^-1
\(\Lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^0\) = 40.95 cm² mol^-1
\(\Lambda_{\mathbf{H}^{+}}^0\) = 349.15 cm² mol^-1
Answer:
The degree of dinociation α is given us
α = \(\frac{\Lambda_{\mathrm{m}}^{\mathrm{c}}}{\Lambda_{\mathrm{m}}^0}\)

Intext Questions – Answers

Question 1.
How would you determine the standard electrode potential of the system Mg²⁺ | Mg?
Answer:
Set up a cell consisting of Mg | MgSO₄ (1M) as one electrode (by dipping a magnesium rod in 1M MgSO₄ solution) and standard hydrogen electrode Pt,H₂(1 atm) | H⁺ (1M) as the second electrode and measure the EMF of the cell. Also note the direction of deflection in the voltmeter. The direction of deflection shows that electrons flow from magnesium electrode to hydrogen electrode, i.e., oxidation takes place at magnesium electrode and reduction at hydrogen electrode. Hence the cell may be represented as follows.
Mg | Mg²⁺ (1M) || H⁺ (1M) | H₂, (1 atm), Pt

Question 2.
Can you store copper sulphate solutions in zinc pot ?
Answer:
As \(\mathrm{E}_{\mathrm{Zn}^{2+} \mid \mathrm{Zn}}\) (-076 V) is lower than
(+0.34 V), therefore, Cu²⁺ ions can oxidise Zn to Zn²⁺ ions according to the reaction.
Cu²⁺ (aq) + Zn (s) → Zn²⁺ (aq) + Cu (s)
Hence, CuSO₄ solution cannot be stored in zinc pot.

Question 3.
Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Answer:
Fe²⁺ → Fe³⁺ + \(\mathrm{e}^{-} \mathrm{E}_{\mathrm{oxid}}^{\ominus}\) = – 0.77 V
Only those substances can oxidise Fe²⁺ to Fe³⁺ which are stronger oxidising agents than 0.77V so that the EMF of the cell reaction is positive. This is so for elements lying below Fe³⁺ | Fe²⁺ in the electrochemical series e.g. F₂, Cl₂, Ag⁺ etc.

Question 4.
Calculate the potential of hydrogen ele-ctrode placed in a solution of pH 10.
Answer:
The reduction half reaction for hydrogen electrode is
2H⁺ (aq) + 2e^– → H₂ (g)
Applying Nernst equation
E = \(\mathrm{E}^{\ominus}\) + \(\frac{0.059}{2}\)log [H⁺]²
= 0 + 0.059 log [H⁺]
pH = 10
∴ – log [H⁺] = -10
(or) log [H⁺] = – 10
∴ E = 0.059 (-10) = -0.59V

Question 5.
Calculate the emf of the cell in which the following reaction is taking place :
Ni (s) + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag(s)
Given that \(\mathbf{E}_{\text {(cell) }}^{\ominus}\) = 1.05 V
Answer:
The cell reaction is
Ni (s) + 2Ag⁺ (aq) → Ni²⁺ (aq) + 2Ag (s)
Here Ni / Ni²⁺ electrode is anode and Ag⁺ / Ag electrode is cathode.
Applying Nernst equation to above system we have

Question 6.
The cell in which the following cell reaction occurs:
2Fe³⁺ (aq) + 2e^– (aq) → 2 Fe²⁺ (aq) + I₂(s) has \(E_{\text {cell }}^{\ominus}\) = 0.236V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Answer:
2Fe³⁺ (aq) + 2e^– (aq) → 2 Fe²⁺ or 2I^– → I₂ + 2e^–
Hence, for the given cell reaction n = 2

Question 7.
Why does the conductivity of a solution decrease with dilution?
Answer:
Conductivity of a solution is the conductance of ions present in a unit volume of the solution. On dilution the number of ions per unit volume decreases. Hence the conductivity decreases.

Question 8.
Suggest a way to determine the \(\Lambda_{\mathrm{m}}^{\circ}\) value of water.
Answer:

Question 9.
The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1. Calcu-late its degree of dissociation and dissoci-ation constant. Given λ⁰ (H⁺) = 349.6 S cm² mol^-1 and λ⁰ (HCOO^–) = 54.6 S cm² mol^-1.
Answer:
\(\Lambda_{\mathrm{HCOOH}}^0\) = \(\lambda_{\mathrm{H}^{+}}^0\) + \(\lambda_{\mathrm{HCOO}^{-}}^0\)
= 349.6 + 54.6 S cm² mol^-1
= 404.2 S cm² mol^-1
\(\Lambda^{\mathrm{c}}\) = 46.1 S cm² mol^-1 (Given)
∴ α = \(\frac{\Lambda^c}{\Lambda^0}\) = \(\frac{46.1}{404.2}\) = 0.114

Question 10.
If a currect of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire.
Answer:
Current strength = 0.5 A
Time = 2 hr = 2 × 60 × 60 = 7200 s
Quantity of electricity = 0.5 × 7200 = 3600 C
Electrons flowing for 96500 C of electricity
= 6.02 × 10²³
Electrons flowing for 3600 C of electricity
= \(\frac{6.02 \times 10^{23} \times 3600}{96500}\) = 6.02 × 10²³

Question 11.
Suggest a list of metals that are extracted electrolytically.
Answer:
Na, K, Ca, Mg and Al (i.e., cations of 1, 2 and 13 groups).

Question 12.
Consider the reaction

What is the quantity of electricity in coulombs needed to reduce 1 mol of \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\)?
Answer:
From the given reaction 1 mol of \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) ions require 6F = 6 × 96500 C = 579000 C of electricity for reduction of Cr³⁺.

Question 13.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging ?
Answer:
During recharging of lead storage battery, electrical energy is supplied externally. The following reactions will take place.

Question 14.
Suggest two materials other than Hydrogen that can be used as fuels in fuel cells.
Answer:
Methane and Methanol

Question 15.
Explain how rusting of Iron is envisaged as setting up of an electrochemical cell.
Answer:
Rusting of Iron occurs in the presence of water and air aiid it may be considered as electrochemical phenomenon.

Oxidation : Fe (s) Fe²⁺ (aq) + 2e^–
Reduction: O₂ (g) + 4H⁺ (aq) + 4e^– → 2H₂O (1)
Atomospheric
oxidation : 2Fe²⁺ (aq) + 2H₂O (1) + 1/2O₂ (g) → Fe₂O₃ (s) + 4H⁺ (aq)
At a particular spot of Iron, oxidation takes place and that spot acts as anode.
Anode : 2Fe (s) → 2Fe⁺² + 4e,

The electrons released at anodic spot go to mother spot on the metal and reduce oxygen at that spot in the presence of H⁺
Cathode : O₂ (g) + \(4 \mathrm{H}^{+} \text {(aq) }\) + 4e → 2H₂O (l) \(\mathrm{E}_{\mathrm{H}^{+}\left|\mathrm{O}_2\right| \mathrm{H}_2 \mathrm{O}}^{\ominus}\) = 1.23 V
Overall reaction :
2Fe(s) + O₂ (g) + 4H⁺ → 2Fe⁺² (aq) + 2H₂O (l), \(\mathrm{E}_{\text {cell }}^{\ominus}\) = 1.67V
The Ferrous ions are further oxidised by atmospheric oxygen to Ferric ions which come out as rust in the form of hydrated Ferric oxide.

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Differential Equations Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2B Differential Equations Important Questions Very Short Answer Type

Question 1.
Find the order and degree of the differential equation \(\frac{d^2 y}{d x^2}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^{5 / 3}\). [(TS) May ’18; May ’13]
Solution:
Given differential equation is \(\frac{d^2 y}{d x^2}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^{5 / 3}\)
Cubing on both sides we get,
\(\left(\frac{d^2 y}{d x^2}\right)^3=\left[1+\left(\frac{d y}{d x}\right)^2\right]^5\)
∴ Order = Order of \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 2
Degree = The degree of \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 3

Question 2.
Find the order and degree of \(1+\left(\frac{d^2 y}{d x^2}\right)^2=\left[2+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}\)
Solution:
Given differential equation is \(\frac{d^2 y}{d x^2}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^{5 / 3}\)
Cubing on both sides,
\(\left(\frac{d^2 y}{d x^2}\right)^3=\left[1+\left(\frac{d y}{d x}\right)^2\right]^5\)
The given equation is a polynomial equation in \(\frac{d^2 y}{d x^2}, \frac{d y}{d x}\)
Hence, \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) is the highest derivative occurring in the equation.
Its order = 2
The degree = 3

Question 3.
Find the order and degree of \(\left[\left(\frac{d y}{d x}\right)^{1 / 2}+\left(\frac{d^2 y}{d x^2}\right)^{1 / 3}\right]^{1 / 4}=0\). [May ’12]
Solution:
Given differential equation is

∴ Order = Order of \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 2
Degree = The degree of \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 2

Question 4.
Find the order and degree of \(\left(\frac{d^3 y}{d x^3}\right)^2-3\left(\frac{d y}{d x}\right)^2-e^x=4\). [May ’14]
Solution:
Given differential equation is \(\left(\frac{d^3 y}{d x^3}\right)^2-3\left(\frac{d y}{d x}\right)^2-e^x=4\)
It is a polynomial equation in \(\frac{d^3 y}{dx^3}, \frac{d y}{d x}\)
Hence, \(\frac{\mathrm{d}^3 \mathrm{y}}{\mathrm{dx}^3}\) is the highest derivative occurring in the equation.
Its order = 3
degree = 2

Question 5.
Find the order and degree of \(x^{1 / 2}\left(\frac{d^2 y}{d x^2}\right)^{1 / 3}+x \frac{d y}{d x}+y=0\). [(AP) Mar. ’18, (TS) ’15]
Solution:

It is a polynomial equation in \(\frac{d^2 y}{d x^2} \& \frac{d y}{d x}\)
Hence \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) is the highest derivative occurring in the equation.
Order = 2
Degree = 1

Question 6.
Find the order and degree of \(\left[\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3\right]^{6 / 5}=6 y\). [(TS) May ’17, ’15; (AP) Mar. ’16, May ’15]
Solution:
Given differential equation is
\(\left[\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^3\right]^{6 / 5}=6 y\)
\(\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3=(6 y)^{5 / 6}\)
∴ Order = Order of \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 2
Degree = The degree of \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 1

Question 7.
Form the differential equation corresponding to y = cx – 2c², where c is a parameter. [Mar. ’19 (AP) ’12]
Solution:
The given equation is y = cx – 2c², where ‘c’ is a parameter. ……..(1)
Differentiating with respect to x on both sides
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = c – 0
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = c ……..(2)
From (1) and (2),
y = \(x\left(\frac{d y}{d x}\right)-2\left(\frac{d y}{d x}\right)^2\)
which is the required differential equation.

Question 8.
Form the differential equation corresponding to y = A cos 3x + B sin 3x, where A and B are parameters. [(TS) Mar. ’20, May ’16; (AP) Mar. ’15; May ’14]
Solution:
The given equation is y = A cos 3x + B sin 3x where A, B are parameters. …….(1)
Differentiate with respect to ‘x’ on both sides
y1 = A (-sin 3x) 3 + B (cos 3x) 3 = -3A sin 3x + 3B cos 3x
Again differentiate with respect to ‘x’ on both sides
y2 = -3A (cos 3x) 3 + 3B (-sin 3x) 3
= -9A cos 3x – 9B sin 3x
= -9 (A cos 3x + B sin 3x)
= -9y (from (1))
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) + 9y = 0 or y² + 9y = 0
which is the required differential equation.

Question 9.
Form the differential equation corresponding to y = a cos(nx + b), where a, b are parameters. [(AP) May ’17]
Solution:
Given y = a cos(nx + b) ……..(1)
(a, b are parameters)
Diff. (1) w.r. to x on both sides we get,
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a[-sin(nx + b) (n)] = -an sin(nx + b) ……….(2)
Again diff. (2) w.r.to x on both sides we get,
\(\frac{d^2 y}{d x^2}\) = -an [cos(nx + b) n]
= -an² cos(nx + b)
= -n²y (from (1))
Required D.E is \(\frac{d^2 y}{d x^2}\) + n2y = 0

Question 10.
Find the order and degree of the differential equation of the family of all circles with their centres at the origin. [(TS) Mar. ’17, ’11]
Solution:
Equation of the family of circles with their centres at the origin is x² + y² = r²
Differentiate with respect to ‘x’ we get,
2x + 2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
⇒ x + y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
∴ Order = Order of \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 1
Degree = The degree of \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 1

Question 11.
Obtain the differential equation which corresponds to the rectangular hyperbolas which have the coordinate axes as asymptotes.
Solution:
The equation of the family of the rectangular hyperbolas which have the coordinate axes as asymptotes are xy = c² …….(1)
where c is a parameter
Diff. (1) w.r. t. x on both sides, we get
x \(\frac{d y}{d x}\) + y = 0
∴ Required D.E is x \(\frac{d y}{d x}\) + y = 0

Question 12.
Find the general solution of x + y \(\frac{d y}{d x}\) = 0. [(AP) May ’16]
Solution:
Given D.E is x + y \(\frac{d y}{d x}\) = 0
x dx + y dy = 0
Integrating on both sides we get
∫x dx + ∫y dy = c
\(\frac{x^2}{2}+\frac{y^2}{2}\) = c
x² + y² = 2c
Which is the required solution of D.E.

Question 13.
Find the general solution of \(\frac{d y}{d x}=\frac{2 y}{x}\). [(TS) Mar. ’19, (AP) ’17]
Solution:
Given D.E. is \(\frac{d y}{d x}=\frac{2 y}{x}\)
⇒ \(\frac{1}{y} d y=\frac{2}{x} d x\)
⇒ \(\int \frac{1}{y} d y=\int \frac{2}{x} d x\)
⇒ log y = 2 log x + log c
⇒ log y = log x² + log c
⇒ log y = log(x²c)
⇒ y = cx² is the required D.E

Question 14.
Find the general solution of \(\frac{\mathbf{d y}}{\mathbf{d x}}\) = e^x+y. [(TS) Mar. ’18]
Solution:
Given \(\frac{\mathbf{d y}}{\mathbf{d x}}\) = e^x+y
⇒ \(\frac{\mathbf{d y}}{\mathbf{d x}}\) = e^x . e^y
⇒ e^-y dy = e^x dx
⇒ ∫e^-y dy = ∫e^x dx
⇒ -e^-y = e^x + c
⇒ e^x + e^-y = c

Question 15.
Find the general solution of \(\sqrt{1-x^2} d y+\sqrt{1-y^2} d x=0\)
Solution:
Given D.E is \(\sqrt{1-x^2} d y+\sqrt{1-y^2} d x=0\)

sin^-1(y) = -sin^-1(x) + c
sin^-1(x) + sin^-1(y) = c
Which is the required general solution.

Question 16.
Solve \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\). [(TS) May ’19]
Solution:
Given differential equation is
\(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
⇒ \(\frac{d y}{1+y^2}=\frac{d x}{1+x^2}\)
Integrating on both sides
\(\int \frac{d y}{1+y^2}=\int \frac{d x}{1+x^2}\)
tan^-1(y) = tan^-1(x) + c
Which is the required solution.

Question 17.
Solve \(\frac{\mathbf{d y}}{\mathbf{d x}}\) + 1 = e^x+y
Solution:

Question 18.
Form the differential equation corresponding to the family of circles passing through the origin and having centres on the Y-axis.
Solution:
The equation of the family of circles passing through the origin and having centres on the Y-axis is x² + y² + 2fy = 0 …….(1)
where f is a parameter.
Diff. (1) w.r. to ‘x’ on both sides, we get

Question 19.
Find the order of the differential equation corresponding to y = Ae^x + Be^3x + Ce^5x. (A, B, C being parameters)
Solution:
Given equation is y = Ae^x + Be^2x + Ce^5x, (A, B, C are parameters)
Here, no. of arbitrary constants = 3
∴ The order of D.E = 3

Question 20.
Form the differential equation corresponding to xy = ae^x + be^-x where a, b are parameters. [(AP) May ’19]
Solution:
Given xy = ae^x + be^-x ……..(1)
(a, b are parameters)
Diff. (1) with respect to x on both sides, we get

Question 21.
Obtain the differential equation which corresponds to the ellipses with centres at the origin and having coordinate axes as axes.
Solution:
The equation of the family of ellipses with centres at the origin and having coordinate axes as axes is
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) ……..(1)
(a, b are arbitrary constants)
Diff. (1) w.r. t. x on both sides, we get

Which is the required solution.

Question 22.
Find the I.F of (cos x) \(\frac{d y}{d x}\) + y sin x = tan x.
Solution:
Given diff. equation is
\(\frac{d y}{d x}\) + y tan x = sec x . tan x
It is a linear diff. equation in y.
Comparing it with \(\frac{d y}{d x}\) + Py = Q,
where P = tan x, Q = sec x tan x
I.F = e^{∫P dx}
= e^{∫tan x dx}
= e^{log|sec x| x}
= sec x

Question 23.
Find I.F of x \(\frac{d y}{d x}\) – y = 2x² sec²2x.
Solution:
Given diff. equation is x \(\frac{d y}{d x}\) – y = 2x² sec²2x
Dividing with x on both sides we get,
\(\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}\) = 2x sec²2x
\(\frac{d y}{d x}+\left(\frac{-1}{x}\right) y\) = 2x sec²2x
It is a linear diff. equation in y of the first order.
Comparing it with \(\frac{d y}{d x}\) + Py = Q
We get P = \(\frac{-1}{x}\), Q = 2x sec²2x

Question 24.
Find the order of the differential equation obtained by eliminating the arbitrary constants b and c from xy = ce^x + be^-x + x².
Solution:
Given equation is xy = ce^x + be^-x + x²
order of the differential equation = number of arbitary constants = 2.

Question 25.
Form the differential equation of the family of curves y = (a + bx) e^kx where a, b are parameters.
Solution:
Given equation is y = (a + bx) e^kx ……..(1)
a, b are arbitrary constants
differentiating (1) w.r.t ‘x’ on both sides, we get

Again differentiating (2) w.r.t ‘x’ on both sides

Which is the required differential equation.

Question 26.
Form the differential equation of the family of curves y = ax² + bx, where a, b are parameters.
Solution:
Given equation is y = ax² + bx ……..(1)
a, b are arbitrary constants
differentiating (1) w.r.t ‘x’ on both sides
\(\frac{d y}{d x}\) = a(2x) + b(1)
\(\frac{d y}{d x}\) = 2ax + b …….(2)
Again differentiating (2) w.r.t ‘x’ on both sides


Which is the required differential equation.

Question 27.
Form the differential equation of the family of curves ax² + by² = 1, where a, b are parameters.
Solution:
Given equation is ax² + by² = 1 …….(1)
a, b are parameters
differentiating (1) w.r.t ‘x’ on both sides

Again differentiating (2) w.r.t ‘x’ on both sides

Which is the required differential equation.

Question 28.
Form the differential equation of the family of curves xy = ax² + \(\frac{b}{x}\), where a, b are parameters.
Solution:
Given equation is xy = ax² + \(\frac{b}{x}\)
xy = \(\frac{a x^3+b}{x}\)
x²y = ax³ + b …….(1)
a, b are parameters
differentiating (1) w.r.t ‘x’ on both sides
x² \(\frac{d y}{d x}\) + y . 2x = a . 3x² + 0
x² \(\frac{d y}{d x}\) + 2xy = 3ax² ……..(2)
Again differentiating (2) w.r.t ‘x’ on both sides

Substitute the value of ‘a’ in (2)

Which is the required D.E.

Question 29.
Find the differential equation of the family of circles that touch the y-axis at the origin.
Solution:
The equation of the family of circles that touch the y-axis at the origin is x² + y² + 2gx = 0 …….(1)
g is a parameter.
differentiating (1) w.r.t ‘x’ on both sides, we get
2x + 2y \(\frac{d y}{d x}\) + 2g = 0
x + y \(\frac{d y}{d x}\) + g = 0
∴ g = -x – y \(\frac{d y}{d x}\)
Substitute the value of ‘g’ in (1)
x² + y² + 2x[-x – y \(\frac{d y}{d x}\)] = 0
x² + y² – 2x² – 2xy \(\frac{d y}{d x}\) = 0
y² – x² – 2xy \(\frac{d y}{d x}\) = 0
Which is the required D.E.

Question 30.
Obtain the D.E. corresponding to the parabolas each of which has a latus rectum 4a and whose axis is parallel to the x-axis.
Solution:
The equation of the family of parabolas each of which has a latus rectum 4a and whose axis is parallel to the x-axis is
(y – k)² = 4a(x – h) ……..(1)
h, k are parameters differentiating (1) w.r.t ‘x’ on both sides, we get
2(y – k) \(\frac{d y}{d x}\) = 4a(1 – 0)
\(\frac{d y}{d x}\) . 2(y – k) = 4a
\(\frac{d y}{d x}\) (y – k) = 2a …….(2)
Again differentiating (2) w.r. t ‘x’ on both sides, we get


Which is the required D.E.

Question 31.
Obtain the D.E. corresponding to the parabolas having their focus at the origin and axis along the x-axis.
Solution:
The equation of the family of parabolas having their foci at the origin and axis along the x-axis is
y² = 4a(x + a) ……..(1), ‘a’ is a parameter
differentiating (1) w.r.t ‘x’ on both sides


Which is the required D.E.

Question 32.
Express the D.E. \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\) of f(x) dx + g(y) dy = 0.
Solution:
Given differential equation is \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
\(\frac{d y}{1+y^2}=\frac{d x}{1+x^2}\)
\(\frac{d x}{1+x^2}-\frac{d y}{1+y^2}=0\)
This is of the form f(x) dx + g(y) dy = 0

Question 33.
Express the D.E. y – x \(\frac{d y}{d x}\) = a(y² + \(\frac{d y}{d x}\)) in the form of f(x) dx + g(y) dy = 0.
Solution:
Given D.E. is y – x \(\frac{d y}{d x}\) = a(y² + \(\frac{d y}{d x}\))

This is of the form f(x) dx + g(y) dy = 0.

Question 34.
Express the D.E. \(\frac{d y}{d x}\) = e^x-y + x² e^-y in the form of f(x) dx + g(y) dy = 0.
Solution:
Given D.E. is \(\frac{d y}{d x}\) = e^x-y + x² e^-y
⇒ \(\frac{d y}{d x}\) = e^x . e^-y + x² e^-y
⇒ \(\frac{d y}{d x}\) = e^-y (e^x + x²)
⇒ \(\frac{d y}{e^{-y}}\) = (e^x + x²) dx
⇒ e^y dy = (e^x + x²) dx
⇒ (e^x + x²) dx – e^y dy = 0
This is of the form f(x) dx + g(y) dy = 0

Question 35.
Express the D.E. \(\frac{d y}{d x}\) + x² = x² e^3y in the form of f(x) dx + g(y) dy = 0.
Solution:
Given D.E. is \(\frac{d y}{d x}\) + x² = x² e^3y
⇒ \(\frac{d y}{d x}\) = x² e^3y – x²
⇒ \(\frac{d y}{d x}\) = x² (e^3y – 1)
⇒ \(\frac{\mathrm{dy}}{\mathrm{e}^{3 \mathrm{y}}-1}\) = x² dx
⇒ x² dx – \(\frac{\mathrm{dy}}{\mathrm{e}^{3 \mathrm{y}}-1}\) = 0
This is of the form f(x) dx + g(y) dy = 0.

Question 36.
Show that f(x, y) = 4x²y + 2xy² is a homogeneous function of degree 3.
Solution:
Given f(x, y) = 4x²y + 2xy²
Now, f(kx, ky) = 4(kx)² (ky) + 2(kx) (ky)²
= 4k² . x² . ky + 2kx . k²y²
= k³ (4x²y + 2xy²)
= k³ f(x, y)
∴ f(x, y) is a homogeneous function of degree 3.

Question 37.
Show that g(x, y) = \(x y^{1 / 2}+y x^{1 / 2}\) is a homogeneous function of degree 3/2.
Solution:
Given g(x, y) = \(x y^{1 / 2}+y x^{1 / 2}\)

∴ g(x, y) is a homogeneous function of degree 3/2.

Question 38.
Show that h(x, y) = \(\frac{x^2+y^2}{x^3+y^3}\) is a homogeneous function of degree = 1.
Solution:
Given h(x, y) = \(\frac{x^2+y^2}{x^3+y^3}\)

∴ h(x, y) is a homogeneous function of degree -1.

Question 39.
Show that f(x, y) = 1 + e^x/y is a homogeneous function of x and y.
Solution:
Given f(x, y) = 1 + e^x/y
Now f(kx, ky) = 1 + \(e^{\left(\frac{k x}{k y}\right)}\)
= 1 + e^x/y
= k⁰ f(x, y)
Hence, f(x, y) is a homogeneous function of degree ‘0’.

Question 40.
Show that f(x, y) = \(x \sqrt{x^2+y^2}-y^2\) is a homogeneous function of x and y.
Solution:
Given f(x, y) = \(x \sqrt{x^2+y^2}-y^2\)

Hence f(x, y) is a homogeneous function of degree 2.

Question 41.
Show that f(x, y) = x – y log y + y log x is a homogeneous function of x and y.
Solution:
Given f(x, y) = x – y log y + y log x
Now f(kx, ky) = kx – ky log (ky) + (ky) log (kx)
= kx – ky (log k + log y) + ky (log k + log x)
= kx – ky log y – ky log k + ky log k + ky log x
= kx – ky log y + ky log x
= k(x – y log y + y log x)
= k f(x, y)
Hence, f(x, y) is a homogeneous function of degree 1.

Question 42.
Express \(\left(1+e^{x / y}\right) d x+e^{x / y}\left(1-\frac{x}{y}\right) d y=0\) in the form of \(\frac{\mathbf{d x}}{\mathbf{d y}}=\mathbf{F}\left(\frac{\mathbf{x}}{\mathbf{y}}\right)\).
Solution:

Which is the required form.

Question 43.
Express \(\frac{\mathbf{d y}}{\mathbf{d x}}=\frac{\mathbf{y}}{x+y \cdot e^{\frac{-2 x}{y}}}\) in the form \(\frac{\mathbf{d x}}{\mathbf{d y}}=\mathbf{F}\left(\frac{x}{\mathbf{y}}\right)\).
Solution:

Which is the required form.

Question 44.
Express x dy – y dx = \(\sqrt{x^2+y^2}\) in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is x dy – y dx = \(\sqrt{x^2+y^2}\) dx

Which is in the required form.

Question 45.
Express \(\left(x-y \tan ^{-1} \frac{y}{x}\right) d x+x \tan ^{-1} \frac{y}{x} d y=0\) in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is

Which is in the required form.

Question 46.
Express x \(\frac{d y}{d x}\) = y(log y – log x + 1) in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is x \(\frac{d y}{d x}\) = y(log y – log x + 1)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}\left[\log \left(\frac{\mathrm{y}}{\mathrm{x}}\right)+1\right]=\mathrm{F}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\)
Which is in the required form.

Question 47.
Transform the D.E. x log x \(\frac{d y}{d x}\) + y = 2 log x.
Solution:
Given D.E. is x log x \(\frac{d y}{d x}\) + y = 2 log x
dividing both sides by x log x, we get
\(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{2 \log x}{x \log x}\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{x \log x}=\frac{2}{\mathrm{x}}\)
Which is in the form \(\frac{d y}{d x}\) + Py = Q
It is linear in y.
Here P = \(\frac{1}{x \log x}\), Q = \(\frac{2}{\mathrm{x}}\)

Question 48.
Transform the D.E. (x + 2y³) \(\frac{\mathrm{dy}}{\mathrm{d} x}\) = y
Solution:
Given D.E. is (x + 2y³) \(\frac{\mathrm{dy}}{\mathrm{d} x}\) = y

Which is in the form \(\frac{\mathrm{dx}}{\mathrm{d} y}\) + Px = Q
It is linear in x.
Here P = \(\frac{-1}{y}\), Q = 2y².

Question 49.
Find the I.F of the D.E. (2x – 10y³) \(\frac{d y}{d x}\) + y = 0 by transforming it into a linear form.
Solution:
Given D.E. is (2x – 10y³) \(\frac{d y}{d x}\) + y = 0
(2x – 10y³) \(\frac{d y}{d x}\) = -y

Question 50.
Find the I.F of the D.E. y . \(\frac{\mathrm{dx}}{\mathrm{dy}}\) – x = 2y³ by transforming it into linear form.
Solution:
Given D.E. is y . \(\frac{\mathrm{dx}}{\mathrm{dy}}\) – x = 2y³
dividing both sides by ‘y’ we get
\(\frac{\mathrm{dx}}{\mathrm{dy}}-\frac{\mathrm{x}}{\mathrm{y}}\) = 2y²
Which is in the form \(\frac{\mathrm{dx}}{\mathrm{dy}}\) + Px = Q
It is linear in x.
Here P = \(\frac{-1}{y}\), Q = 2y²

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Partial Fractions Important Questions to help strengthen their preparations for exams.

TS Inter Second Year Maths 2A Partial Fractions Important Questions

Question 1.
Resolve \(\frac{5 x+1}{(x+2)(x-1)}\) into partial fractions. [May 2000]
Solution:
Let \(\frac{5 x+1}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\)
= \(\frac{A(x-1)+B(x+2)}{(x+2)(x-1)}\)
⇒ 5x + 1 = A (x – 1) + B (x + 2)
Put x = – 2 then 5 (- 2) + 1 = A (- 2 – 1)
⇒ – 3A = – 9
⇒ A = 3
Put x = 1 then 5(1) + I = B(1 + 2)
⇒ 3B = 6
⇒ B = 2
∴ \(\frac{5 x+1}{(x+2)(x-1)}=\frac{3}{x+2}+\frac{2}{x-1}\).

Question 2.
Resolve \(\frac{3 x+7}{x^2-3 x+2}\) into partial fractions. [May ’14]
Solution:
\(\frac{3 x+7}{x^2-3 x+2}=\frac{3 x+7}{(x-1)(x-2)}\)
Let \(\frac{3 x+7}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}\)
= \(\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}\)
3x + 7 = A (x – 2) + B (x – 1)
Put x = 1 then 3(1) + 7 = A (1 – 2)
⇒ 3 + 7 = A (- 1)
⇒ 10 = – A
⇒ A = – 10
Put x = 2 then 3(2) + 7 = B(2 – 1)
⇒ 6 + 7 = B(1)
⇒ B = 13
∴ \(\frac{3 x+7}{(x-1)(x-2)}=\frac{-10}{x-1}+\frac{13}{x-2}\).

Question 3.
Resolve \(\frac{x+4}{\left(x^2-4\right)(x+1)}\) into partial fractions. [Mar. ‘14]
Solution:

Question 4.
Resolve \(\frac{x^2+13 x+15}{(2 x+3)(x+3)^2}\) into partial fractions. [March 2008].
Solution:

Put x = – 3
⇒ C (- 6 + 3) = 9 – 39 + 15
⇒ C(- 3) = – 15
⇒ C = 5
Now, comparing the coefficients of x² in equation (1), we get
A + 2B = 1
⇒ – 1 + 2B = 1
⇒ 2B = 2
⇒ B = 1
∴ \(\frac{x^2+13 x+15}{(2 x+3)(x+3)^2}=\frac{-1}{2 x+3}+\frac{1}{x+3}+\frac{5}{(x+3)^2}\).

Question 5.
Resolve \(\frac{1}{(x-1)^2(x-2)}\) into partial fractions. [May 2013].
Solution:
Let \(\frac{1}{(x-1)^2(x-2)}\) = \(\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x-2)}\)
⇒ \(\frac{1}{(x-1)^2(x-2)}\) = \(\frac{A(x-1)(x-2)+B(x-2)+C(x-1)^2}{(x-1)^2(x-2)}\)
⇒ A (x – 1) (x – 2) + B (x – 2) + C (x – 1)² = 1 …………….(1)
If x = 1
⇒ B(- 1) = 1
⇒ B = – 1
If x = 2
⇒ C(1) = 1
⇒ C = 1
Now, comparing the coefficients of x² in equation (1) we get,
A + C = 0
⇒ A = – 1
∴ \(\frac{1}{(x-1)^2(x-2)}=\frac{(-1)}{x-1}+\frac{(-1)}{(x-1)^2}+\frac{1}{x-2}\).

Question 6.
Resolve \(\frac{2 x^2+2 x+1}{x^3+x^2}\) into partial fractions. [TS – MAr. 2017]
Solution:
\(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{2 x^2+2 x+1}{x^2(x+1)}\)
Let \(\frac{2 x^2+2 x+1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\)
\(\frac{2 x^2+2 x+1}{x^2(x+1)}=\frac{A x(x+1)+B(x+1)+C\left(x^2\right)}{x^2(x+1)}\)
⇒ Ax (x + 1) + B (x + 1) + Cx² = 2x² + 2x + 1 …………….(1)
If x = 0
⇒ B(1) = 1
⇒ B = 1
If x = – 1
⇒ C(- 1)² = 2(1)² – 2 + 1
⇒ C = 1.
Now, comparing the coeff. of x² on both sides of (1) we get,
A+ C = 2
⇒ A + 1 = 2
⇒ A = 1
∴ \(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\).

Question 7.
Resolve \(\frac{3 x-18}{x^3(x+3)}\) into partial fractions.
Solution:
Let \(\frac{3 x-18}{x^3(x+3)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+3}\)
\(\frac{3 x-18}{x^3(x+3)}\) = \(\frac{\mathrm{Ax}^2(\mathrm{x}+3)+\mathrm{Bx}(\mathrm{x}+3)+\mathrm{C}(\mathrm{x}+3)+\mathrm{Dx}^3}{\mathrm{x}^3(\mathrm{x}+3)}\)
⇒ Ax² (x + 3) + Bx (x + 3) + C (x + 3) + D(x³) = 3x – 18 ………………(1)
If x = 0
⇒ C(3) = – 18
⇒ C = – 6
If x = – 3
⇒ D (- 27) = – 9 – 18
⇒ D = 1
Now, comparing the coeff. of x³ in (1) we get,
A + D = 0
⇒ A + 1 = 0
⇒ A = – 1
Now, comparing the coeff. of x² in (1) we get,
3A + B = 0
⇒ 3(- 1) + B = 0,
⇒ B = 3
∴ \(\frac{3 x-18}{x^3(x+3)}=\frac{(-1)}{x}+\frac{3}{x^2}+\frac{(-6)}{x^3}+\frac{1}{x+3}\).

Question 8.
Resolve \(\frac{3 x^2-8 x^2+10}{(x-1)^4}\) into partial fractions. [March 2006]
Solution:
Let x – 1 = y
⇒ x = y + 1

Question 9.
Resolve \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\) into partial fractions. [March ’11, 2007, June ’11, TS – Mar. ’18, AP – MAr. 2015; MAy ’12, ’01]
Solution:
Let \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+2}\)
\(\frac{\mathrm{A}\left(\mathrm{x}^2+2\right)+(\mathrm{Bx}+\mathrm{C})(\mathrm{x}-1)}{(\mathrm{x}-1)\left(\mathrm{x}^2+2\right)}=\frac{2 \mathrm{x}^2+3 \mathrm{x}+4}{(\mathrm{x}-1)\left(\mathrm{x}^2+2\right)}\)
A(x² + 2) + (Bx + C) (x – 1) = 2x² + 3x + 4 ……………….(1)
Put x = 1
⇒ A (1 + 2) = 2 + 3 + 4
⇒ A(3) = 9
⇒ A = 3.
From (1),
⇒ 2x² + 3x + 4 = Ax² + 2A + Bx² – Bx + Cx – C
Now, comparing the coefficients of x2 we get,
A + B = 2
⇒ 3 + B = 2
⇒ B = – 1
Comparing coefficients of x we get,
⇒ – B + C = 3
⇒ 1 + C = 3
⇒ C = 2
∴ \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{3}{x-1}+\frac{(-1) x+2}{x^2+2}\)
= \(\frac{3}{x-1}+\frac{-x+2}{x^2+2}\).

Question 10.
Resolve \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\) into partial fractions. [May ‘09, ‘07, Mar. ‘12, ‘09, ‘05]
[AP – Mar. ‘17, ’16. May ’16; Board Paper].
Solution:
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^2+1}\)
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{A\left(x^2+1\right)+(B x+C)(x+2)}{(x+2)\left(x^2+1\right)}\)
A (x² + 1) + (Bx + C) (x + 2) = x – 3 ……………….(1)
If x = – 2
⇒ A (4 + 1) = 4 – 3
⇒ A = \(\frac{1}{5}\)
From (1),
⇒ Ax² + A + Bx² + 2Bx + Cx + 2C = x² – 3
Now, comparing coeff. of x² on both sides of (1) we get,
A + B = 1
⇒ \(\frac{1}{5}\) + B = 1
⇒ B = 1 – \(\frac{1}{5}\)
⇒ B = \(\frac{4}{5}\)
Comparing coeff. of x on both sides of (1) we get
2B + C = 0
⇒ \(\frac{8}{5}\) + C = O
⇒ C = \(\frac{-8}{5}\)
∴ \(\frac{x^2-3}{(z+2)\left(x^2+1\right)}=\frac{1}{5(x+2)}+\frac{4 x-8}{5\left(x^2+1\right)}\).

Question 11.
Resolve \(\frac{2 x^2+1}{x^3-1}\) into partial fractions.
Solution:
\(\frac{2 x^2+1}{x^3-1}=\frac{2 x^2+1}{(x-1)\left(x^2+x+1\right)}\)
Let \(\frac{2 x^2+1}{(x-1)\left(x^2+x+1\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+x+1}\)
\(\frac{2 x^2+1}{(x-1)\left(x^2+x+1\right)}=\frac{A\left(x^2+x+1\right)+(B x+C)(x-1)}{(x-1)\left(x^2+x+1\right)}\)
A (x² + x + 1) + (Bx + C) (x – 1) = 2x² + 1 ……………….(1)
If x = 1
⇒ A (3) = 3
⇒ A = 1
From (1),
⇒ Ax² + Ax + A + Bx² – Bx + Cx – C = 2x² + 1
Now comparing coeff. of x² on both sides of (1) we get,
A + B = 2
⇒ 1 + B = 2
⇒ B = 1
Now, comparing coeff. of x on both sides of (1) we get, —-
A – B + C = 0
⇒ 1 – 1 + C = 0
⇒ C = 0
∴ \(\frac{2 x^2+1}{x^3-1}=\frac{1}{x-1}+\frac{x}{x^2+x+1}\).

Question 12.
Resolve \(\frac{\mathbf{x}^3}{(\mathbf{x}-\mathbf{a})(\mathbf{x}-\mathbf{b})(\mathbf{x}-\mathbf{c})}\) into partial fractions. [TS – MAy ’16, ’15, May ’08] [TS – Mar. 2019]
Solution:

Question 13.
Resolve \(\frac{x^3}{(2 x-1)(x+2)(x-3)}\) into partial fractions.
Solution:
Let \(\frac{x^3}{(2 x-1)(x+2)(x-3)}\)
= \(\frac{1}{2}+\frac{A}{2 x-1}+\frac{B}{x+2}+\frac{C}{x-3}\)
= \(\frac{x^3}{(2 x-1)(x+2)(x-3)}\)
= \(\frac{(2 \mathrm{x}-1)(\mathrm{x}+2)(\mathrm{x}-3)+2 \mathrm{~A}(\mathrm{x}+2)(\mathrm{x}-3)+2 B(2 \mathrm{x}-1)(\mathrm{x}-3)+2 \mathrm{C}(2 \mathrm{x}-1)(\mathrm{x}+2)}{2(2 \mathrm{x}-1(\mathrm{x}+2)(\mathrm{x}-3)}\)
2x³ = (2x – 1) (x + 2) (x -3) + 2A (x + 2) (x – 3) + 2B (2x – 1) (x – 3) + 2C (2x – 1) (x + 2)
If x = \(\frac{1}{2}\)
⇒ 2A(\(\frac{1}{2}\) + 2) (\(\frac{1}{2}\) – 3) = 2 . \(\frac{1}{8}\)
⇒ 2A . \(\frac{5}{2}\) . \(\frac{-5}{2}\) = \(\frac{1}{4}\)
A = \(-\frac{1}{50}\)
If x = – 2
⇒ 2B (2 (- 2) – 1) (- 2 – 3) = – 2 (8)
⇒ 2B (- 5) (- 5) = – 16
B = \(\frac{-8}{25}\)
If x = 3
⇒ 2C (6 – 1) (3 + 2) = 2 (27)
⇒ 2C (5) (5) = 2(27)
C = \(\frac{27}{25}\)
∴ \(\frac{x^3}{(2 x-1)(x+2)(x-3)}\) = \(\frac{1}{2}+\frac{\frac{-1}{50}}{2 x-1}+\frac{\left(\frac{-8}{25}\right)}{x+2}+\frac{\frac{27}{25}}{x-3}\)
= \(\frac{1}{2}-\frac{1}{50(2 x-1)}-\frac{8}{25(x+2)}+\frac{27}{25(x-3)}\).

Question 14.
Resolve \(\frac{x^4}{(x-1)(x-2)}\) into partial fractions. [TS – Mar. ’16, Mar. ’13, ’10]
Solution:

15x – 14 = A (x – 2) + B (x – 1)
If x = 1
⇒ A(1 – 2) = 15 – 14
⇒ A = – 1
If x = 2
⇒ B(2 – 1) = 15 (2) – 14
⇒ B = 30 – 14
⇒ B = 16
∴ \(\frac{x^4}{(x-1)(x-2)}\) = x² + 3x + 7 – \(\frac{1}{(x-1)}+\frac{16}{(x-2)}\).

Question 15.
Find the coefficient of x⁴ in the expansion \(\frac{3 x}{(x-2)(x+1)}\) of in powers of x. [Mar. ’89, ’97]
Solution:
Let \(\frac{3 x}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}\)
\(\frac{3 x}{(x-2)(x+1)}=\frac{A(x+1)+B(x-2)}{(x-2)(x+1)}\)
⇒ A(x + 1) + B(x – 2) = 3x
If x = 2
⇒ A(3) = 6
⇒ A = 2
If x = – 1
⇒ B (- 3) = -3
⇒ B = 1

Question 16.
Find the coefficient of xⁿ in the power series expansion of \(\frac{x-4}{x^2-5 x+6}\) specfying the region in which the expansion is valid. [May 2010]
Solution:
Given, \(\frac{x-4}{x^2-5 x+6}=\frac{x-4}{(x-3)(x-2)}\)
Let \(\frac{x-4}{(x-3)(x-2)}=\frac{A}{(x-3)}+\frac{B}{(x-2)}\)
\(\frac{\mathrm{x}-4}{(\mathrm{x}-3)(\mathrm{x}-2)}=\frac{\mathrm{A}(\mathrm{x}-2)+\mathrm{B}(\mathrm{x}-3)}{(\mathrm{x}-3)(\mathrm{x}-2)}\)
⇒ A(x – 2) + B (x – 3) = x – 4
If x = 3
⇒ A(1) = – 1
⇒ A = – 1
If x = 2
⇒ B (- 1) = – 2
⇒ B = 2
∴ \(\frac{x-4}{(x-3)(x-2)}=\frac{(-1)}{x-3}+\frac{2}{x-2}\)
Now, \(\frac{-1}{x-3}+\frac{2}{x-2}=\frac{-1}{-3\left(1-\frac{x}{3}\right)}+\frac{2}{(-2)\left(1-\frac{x}{2}\right)}\)
= \(\frac{1}{3}\) (1 – \(\frac{x}{3}\))^-1 – (1 – \(\frac{x}{2}\))^-1 ………….(1)
Now, (1 – \(\frac{x}{3}\))^-1 expansion is valid when
|\(\frac{x}{3}\)| < 1
⇒ |x| < 3
(1 – \(\frac{x}{3}\))^-1 expansion is valid when
|\(\frac{x}{2}\)| < 1
⇒ |x| < 2
Thus, if |x| < 2, both the above expansions are valid.
Hence, we get from (1)
\(\frac{-1}{x-3}+\frac{2}{x-2}\) = \(\frac{1}{3}\left[1+\frac{x}{3}+\left(\frac{x}{3}\right)^2+\ldots \ldots+\left(\frac{x}{3}\right)^n+\ldots \ldots\right]\) – \(\left[1+\frac{\mathrm{x}}{2}+\left(\frac{\mathrm{x}}{2}\right)^2+\ldots \ldots\left(\frac{\mathrm{x}}{2}\right)^{\mathrm{n}}+\ldots .\right]\)
∴ Coeff. of xⁿ = \(\frac{1}{3 \cdot 3^n}-\frac{1}{2^n}=\frac{1}{3^{n+1}}-\frac{1}{2^n}\).

Some More Maths 2A Partial Fractions Important Questions

Question 1.
Resolve \(\frac{2 x+3}{5(x+2)(2 x+1)}\) into partial fractions.
Solution:

Question 2.
Resolve \(\frac{13 x+43}{2 x^2+17 x+30}\) into partial fractions.
Solution:
We have 2x² + 17x + 30 = (2x + 5) (x + 6)
∴ \(\frac{13 x+43}{2 x^2+17 x+30}=\frac{13 x+43}{(2 x+5)(x+6)}\)
Now, let
\(\frac{13 x+43}{(2 x+5)(x+6)}=\frac{A}{2 x+5}+\frac{B}{x+6}\)
∴ 13x + 43 = A(x + 6) + B(2x + 5)
= (A + 2B)x + (6A + 5B)
Comparing the coefficients of like powers of x, we have
A + 2B = 13 and 6A + 5B = 43
Solving these two equations,
we get A = 3 and B = 5
∴ \(\frac{13 x+43}{2 x^2+17 x+30}=\frac{3}{2 x+5}+\frac{5}{x+6}\).

Question 3.
Resolve \(\frac{x^2-x+1}{(x+1)(x-1)^2}\) into partial fractions.
Solution:
Let \(\frac{x^2-x+1}{(x+1)(x-1)^2}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{(x-1)^2}\)
\(\frac{x^2-x+1}{(x+1)(x-1)^2}=\frac{A(x-1)^2+B(x+1)(x-1)+C(x+1)}{(x+1)(x-1)^2}\)
⇒ A(x – 1)² + B(x + 1) (x – 1) + C(x + 1) = x2 – x + 1 …………….(1)
If x = – 1
⇒ A(4) = 3
⇒ A = \(\frac{3}{4}\)
If x = 1
⇒ C(2) = 1
⇒ C = \(\frac{1}{2}\)
Now, comparing the coeff. of x² in equation (1)
We get A + B = 1
⇒ \(\frac{3}{4}\) + B = 1
⇒ B = 1 – \(\frac{3}{4}\)
⇒ B = \(\frac{1}{4}\)
∴ \(\frac{x^2-x+1}{(x+1)(x-1)^2}=\frac{3}{4(x+1)}+\frac{1}{4(x-1)}+\frac{1}{2(x-1)^2}\).

Question 4.
Resolve \(\frac{x-1}{(x+1)(x-2)^2}\) into partial fractions.
Solution:

Question 5.
Resolve \(\frac{1}{x^3(x+a)}\) into partial fractions.
Solution:
Let \(\frac{1}{x^3(x+a)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+a}\)
\(\frac{1}{x^3(x+a)}=\frac{\mathrm{Ax}^2(\mathrm{x}+\mathrm{a})+\mathrm{Bx}(\mathrm{x}+\mathrm{a})+\mathrm{C}(\mathrm{x}+\mathrm{a})+\mathrm{D}\left(\mathrm{x}^3\right)}{\mathrm{x}^3(\mathrm{x}+\mathrm{a})}\)
⇒ Ax² (x + a) + Bx (x + a) + C (x + a) + D(x³) = 1 ……………..(1)
If x = 0
⇒ C(a) = 1
⇒ C = \(\frac{1}{a}\)
If x = – a = 1
⇒ D(- a³) = 1
⇒ D = \(\frac{-1}{a^3}\)
Now, comparing the coeff. of x³ in (1) we get
A + D = 0
⇒ A – \(\frac{1}{a^3}\) = 0
⇒ A = \(\frac{1}{a^3}\)
Now, comparing the coeff. of x² in (1) we get
Aa + B = 0
⇒ a \(\frac{1}{a^3}\) + B = 0
⇒ B = \(-\frac{1}{a^2}\)
∴ \(\frac{1}{x^3(x+a)}=\frac{\frac{1}{a^3}}{x}+\frac{\frac{-1}{a^2}}{x^2}+\frac{\frac{1}{a}}{x^3}+\frac{\frac{-1}{a^3}}{x+a}\)
= \(\frac{1}{a^3 x}+\frac{(-1)}{a^2 x^2}+\frac{1}{a^3}+\frac{(-1)}{a^3(x+a)}\)

Question 6.
Resolve \(\frac{x^2+5 x+7}{(x-3)^3}\) into partial fractions.
Solution:
Let y = x – 3
⇒ x = y + 3
Now,

Question 7.
Resolve \(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\) into partial fractions.
Solution:

If x = 1
⇒ B(3) = 3
⇒ B = 1.
From (1)
⇒ Ax³ – A + Bx² + Bx + B + Cx³ + Cx – 2Cx² + Dx² + D – 2Dx = x³ + x² + 1
Now, comparing coeffi. of x³ we get,
A + C = 1 ……………….(2)
Comparing coeff. of x² we get,
B – 2C + D = 1
⇒ 1 – 2C + D = 1
⇒ – 2C + D = 0 ……………(3)
Comparing coeff. of x we get,
B + C – 2D = 0
⇒ C – 2D = – 1 …………….(4)
Comparing constants we get,
– A + B + D = 1
⇒ – A + D = 0 …………….(5)
Solving (3) and (4)

From (5)
⇒ – A + D = 0
⇒ A + \(\frac{2}{3}\) = 0
⇒ A = \(\frac{2}{3}\)
∴ \(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}=\frac{2}{3(x-1)}+\frac{1}{(x-1)^2}+\frac{x+2}{3\left(x^2+x+1\right)}\).

Question 8.
Resolve \(\frac{x^2}{(x-1)(x-2)}\) into partial fractions.
Solution:
Let \(\frac{x^2}{(x-1)(x-2)}=1+\frac{A}{x-1}+\frac{B}{x-2}\)
\(\frac{x^2}{(x-1)(x-2)}=\frac{(x-1)(x-2)+A(x-2)+B(x-1)}{(x-1)(x-2)}\)
x² = (x – 1) (x – 2) + A (x – 2) B (x – 1)
If x = 1
⇒ A (1 – 2) = 1
⇒ A (- 1) = 1
⇒ A = – 1
If x = 2
⇒ B (2 – 1) = 4
⇒ B (1) = 4
⇒ B = 4
∴ \(\frac{x^2}{(x-1)(x-2)}=1+\frac{(-1)}{x-1}+\frac{4}{x-2}\)
= 1 – \(\frac{1}{(x-1)}+\frac{4}{(x-2)}\).

Question 9.
Resolve \(\frac{x^3}{(2 x-1)(x-1)^2}\) into partial fractions.
Solution:
Let \(\frac{x^3}{(2 x-1)(x-1)^2}\) = \(\frac{1}{2}+\frac{A}{2 x-1}+\frac{B}{(x-1)}+\frac{C}{(x-1)^2}\)
⇒ \(\frac{x^3}{(2 x-1)(x-1)^2}\) = \(\frac{(2 \mathrm{x}-1)(\mathrm{x}-1)^2+\mathrm{A}(2)(\mathrm{x}-1)^2+2 B(2 x-1)(x-1)+2 C(2 x-1)}{2(2 x-1)(x-1)^2}\)
⇒ 2x³ = (2x – 1) (x – 1)² + 2A (x – 1)² + 2B(2x – 1) (x – 1) + 2C (2x – 1)
If x = \(\frac{1}{2}\)
⇒ 2A (\(\frac{1}{2}\) – 1)² = 2(\(\frac{1}{2}\))³
⇒ 2A (\(\frac{1}{4}\)) = 2 . \(\frac{1}{8}\)
⇒ A =\(\frac{1}{2}\)
If x = 1
⇒ 2C (2 – 1) = 2
⇒ C = i
Comparing the coefficients of x², we get
2A + 4B – 5 = 0
2 (\(\frac{1}{2}\)) + 4B – 5 = 0
⇒ 4B = 4
⇒ B = 1.
∴ \(\frac{x^3}{(2 x-1)(x-1)^2}=\frac{1}{2}+\frac{\frac{1}{2}}{2 x-1}+\frac{(+1)}{x-1}+\frac{1}{(x-1)^2}\)
= \(\frac{1}{2}+\frac{1}{2(2 x-1)}+\frac{1}{x-1}+\frac{1}{(x-1)^2}\).

Question 10.
Resolve \(\frac{x^3}{(x-1)(x+2)}\) into partial fractions. [Ap – Mar. 2019]
Solution:
Let (x – 1) (x + 2) = x² + 2x – x – 2 = x

A(x + 2) + B(x – 1) = 3x – 2
If x = 1
⇒ A (1 + 2) = 3 – 2
⇒ A = \(\frac{1}{3}\)
If x = – 2
⇒ B (- 2 – 1) = 3 (- 2) – 2
B (- 3) = – 8
⇒ B = \(\frac{8}{3}\)
∴ \(\frac{x^3}{(x-1)(x+2)}\) = x – 1 + \(\frac{\frac{1}{3}}{x-1}+\frac{\frac{8}{3}}{x+2}\)
= x – 1 + \(\frac{1}{3(x-1)}+\frac{8}{3(x+2)}\).

Question 11.
Find the coefficient of x³ in the power series expansion of \(\frac{5 x+6}{(x+2)(1-x)}\) specifying the region in which the expansion is valid.
Solution:

Question 12.
Resolve \(\frac{x^3+x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}\) into partial fractions.
Solution:
Let \(\frac{x^3+x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{A x+B}{x^2+2}+\frac{C x+D}{x^2+3}\)
\(\frac{x^3+x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}\) = \(\frac{(\mathrm{Ax}+\mathrm{B})\left(\mathrm{x}^2+3\right)+(\mathrm{Cx}+\mathrm{D})\left(\mathrm{x}^2+2\right)}{\left(\mathrm{x}^2+2\right)\left(\mathrm{x}^2+3\right)}\)
(Ax + B) (x² + 3) + (Cx + D) (x² + 2) = x³ + x² + 1
Ax³ 3Ax + Bx² + 3B + Cx³ + 2Cx + Dx² + 2D = x³ + x² + 1
Now, comparing the coefficients of x3 we get,
⇒ A + C = 1 ………………(1)
Comparing the coefficients of x2 we get,
⇒ B + D = 1 ………………(2)
Comparing the coefficients of x, we get,
⇒ 3A + 2C = 0 ……………..(3)
Comparing the constant terms we get,
⇒ 3B + 2D = 1 …………….(4)
Solve (1) and (3)

Question 13.
Resolve \(\frac{3 x^3-2 x^2-1}{x^4+x^2+1}\) into partial fractions.
Solution:
Let \(\frac{3 x^3-2 x^2-1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\) = x⁴ + 2x² + 1 – x²
= (x² + 1)² – x²
= (x² + x + 1) (x² – x + 1)
= \(\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1}\)
\(\frac{3 x^3-2 x^2-1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\) = \(\frac{(A x+B)\left(x^2-x+1\right)+(C x+D)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
(Ax + B) (x² – x + 1) + (Cx + D) (x² + x + 1) = 3x³ – 2x² – 1
AX³ – Ax² + Ax + Bx² – Bx + B + Cx³ + Cx² + Cx + Dx² + Dx + D = 3x³ – 2x² – 1
Now, comparing the coeff. of x³ on both sides we get
⇒ A + C = 3 …………….(1)
Comparing the coefi. of x² on both sides,
we get
⇒ – A + B + C + D = – 2 ……………..(2)
Comparing the coeff. of xon both sides, we get
A – B + C + D = 0
⇒ 3 – B + D = 0
⇒ B – D = 3 ………………..(3)
Comparing constants on both sides we get,
B + D = – 1 ……………(4)
From (3) and (4),
⇒ B – D = 3
B + D = – 1
2B = 2
B = 1
from B – D = 3
1 – D = 3
D = – 2
From (2)
⇒ – A + B + C + D = – 2
⇒ – A + 1 + C – 2 = – 2
– A + C = – 1 ……………(5)
Solving (1) and (5)
A + C = 3
– A + C = – 1
2C = + 2
C = + 1
from A + C = 3
A + 1 = 3
A = 2
∴ \(\frac{3 x^3-2 x^2-1}{x^4+x^2+1}=\frac{2 x+1}{x^2+x+1}+\frac{x-2}{x^2-x+1}\).

Question 14.
Resolve \(\frac{x+3}{(1-x)^2\left(1+x^2\right)}\) into partial fractions.
Solution:
Let \(\frac{x+3}{(1-x)^2\left(1+x^2\right)}\) = \(\frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C x+D}{1+x^2}\)
\(\frac{x+3}{(1-x)^2\left(1+x^2\right)}\) = \(\frac{A(1-x)\left(1+x^2\right)+B\left(1+x^2\right)+(C x+D)(1-x)^2}{(1-x)^2\left(1+x^2\right)}\)
⇒ A (1 – x) (1 + x²) + B (1 + x²) + (Cx + D) (1 – x)² = x + 3 ………………(1)
If x = 1
⇒ B(1 + 1) = 1 + 3
⇒ B(2) = 4
⇒ B = 2.
From (1)
⇒ A(1 + x² – x – x³) + B (1 + x²) + (Cx + D) (1 + x² – 2x) = x + 3
⇒ A + Ax² – Ax – Ax³ + B + Bx² + Cx + Cx³ – 2Cx² + D + Dx² – 2Dx = x + 3
Now, comparing the coefficients of x³ on both sides we get,
⇒ – A + C = 0
⇒ A – C = 0 ……………..(2)
Corn paring the coefficients of x² we get,
A + B – 2C + D = 0
A + 2 – 2C + D = 0
⇒ A – 2C + D = – 2 ………………(3)
Comparing coefficients of x we get,
– A + C – 2D = 1
⇒ 0 – 2D = 1
⇒ D = \(-\frac{1}{2}\)
Comparing constants we get,
A + B + D = 3 ………………(4)
From (4)
⇒ A + 2 – \(\frac{1}{2}\) = 3
⇒ A = \(\frac{3}{2}\)
From (2)
⇒ \(\frac{3}{2}\) – C = 0
C = \(\frac{3}{2}\)
∴ \(\frac{x+3}{(1-x)^2\left(1+x^2\right)}\) = \(\frac{\frac{3}{2}}{(1-x)}+\frac{2}{(1-x)^2}+\frac{\frac{3}{2} x-\frac{1}{2}}{1+x^2}\)
= \(\frac{3}{2(1-x)}+\frac{2}{(1-x)^2}+\frac{3 x-1}{2\left(1+x^2\right)}\).

Question 15.
Resolve \(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\) into partial fractions.
Solution:
Let \(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\) = \(\frac{A x+B}{1-x+x^2}+\frac{C}{x+2}\)
\(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}=\frac{(A x+B)(x+2)+C\left(1-x+x^2\right)}{\left(1-x+x^2\right)(x+2)}\)
(Ax + B) (x + 2) + C (1 – x + x²) = 3x – 1 …………….(1)
If x = – 2
⇒ C (1 + 2 + 4) = – 6 – 1
⇒ C(7) = – 7
⇒ C = – 1
From (1)
⇒ Ax² + 2Ax+ Bx + 2B + C – Cx + Cx² = 3x – 1
Now. comparing coeff. of x² we get,
⇒ A + C = 0
⇒ A – 1 = 0
⇒ A = 1
Comparing coeff. of x we get,
2A + B – C = 3
⇒ 2 + B + 1 = 3
⇒ B + 3 = 3
⇒ B = 0
∴ \(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}=\frac{x}{1-x+x^2}+\frac{(-1)}{x+2}\).

Question 16.
Resolve \(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\) into partial fractions.
Solution:

A (x – 1) (x² + x + 1) + B (x² + x + 1) + (Cx + D) (x – 1)² = x³ + x² + 1 ………………(1)
II x = 1
⇒ B(3) = 3
⇒ B = 1
From (1)
⇒ Ax³ + Ax² + Ax – Ax² – Ax – A + Bx² + Bx + B + Cx³ + Cx – 2Cx² + Dx² + D – 2Dx = x³ + x² + 1
Now, comparing coefficients of x³ we get,
A + C = 1 ………….(2)
Comparing coeff. of x² we get,
B – 2C + D = 1
⇒ 1 – 2C + D = 1
⇒ – 2C + D = 0 …………….(3)
Comparing coeff. of x we get,
B + C – 2D = 0
⇒ C – 2D = – 1
Comparing constants we get,
⇒ – A + B + D = 1
⇒ – A + D = 0 ………….(5)
Solving (3) and (4)

Question 17.
Resolve \(\frac{x^3}{(2 x-1)(x-1)^2}\) into partial fractions.
Solution:

If x = 1
⇒ 2C (2 – 1) = 2
⇒ C = 1
Comparing the coefficients of x we get,
2A + 4B – 5 = 0
2(\(\frac{1}{2}\)) + 4B – 5 = o
⇒ B = 1
∴ \(\frac{x^3}{(2 x-1)(x-1)^2}=\frac{1}{2}+\frac{\frac{1}{2}}{2 x-1}+\frac{(+1)}{x-1}+\frac{1}{(x-1)^2}\)
= \(\frac{1}{2}+\frac{1}{2(2 x-1)}+\frac{1}{x-1}+\frac{1}{(x-1)^2}\)

Question 18.
Find the coefficient of xⁿ in the power series expansion of \(\frac{x}{(x-1)^2(x-2)}\) specifying the region In which the expansion is valid. [AP – May 2015].
Solution:
Let \(\frac{x}{(x-1)^2(x-2)}\) = \(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x-2}\)
\(\frac{x}{(x-1)^2(x-2)}\) = \(\frac{A(x-1)(x-2)+B(x-2)+C(x-1)^2}{(x-1)^2(x-2)}\)
A(x – 1) (x – 2) + B (x – 2) + C(x – 1)² = x ……………….(1)
If x = 1
⇒ B (- 1) = 1
⇒ B = – 1
If x = 2
⇒ C(1) = 2
⇒ C = 2
Now comparing coefficients of x² on both sides of (1)
A + C = 0
⇒ A + 2 = 0
⇒ A = – 2
∴ \(\frac{x}{(x-1)^2(x-2)}=\frac{(-2)}{x-1}+\frac{(-1)}{(x-1)^2}+\frac{2}{(x-2)}\).

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements

Very Short Answer Questions (2 Marks)

Question 1.
What inspired Bartlett for carrying out reaction between Xe and PtF₆?
Answer:
Neil Bartlett, first prepared a red compound which is formulated as Q₂⁺ PtF₆^–. He, then realised that the first ionisation enthalpy of molecular oxygen was almost identical with that of Xenon. He then made another red colour compound XePtF₆ by mixing PtF₆ and Xenon.

Question 2.
Which of the following does not exist?
a) XeOF₄
b) NeF₂
c) XeF₂
d) XeF₆
Answer:
NeF₂ does not exist.

Question 3.
Why do noble gases have comparatively large atomic sizes?
Answer:
The atomic radii of the noble gas elements are all very large. Because these have van der Waals radii which Is larger than Ionic and covalent radii.

Question 4.
List out the uses of neons. [Mar. 18 – A.P.]
Answer:

1. Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes.
2. Neon bulbs are used in botanical gardens and in green houses.

Question 5.
Write any two uses of argon.
Answer:

1. Argon is used to provide an inert atmosphere in high temperature metallurgical processes.
2. For filling electric bulbs.
3. Used in laboratory for handling substances that are air sensitive.

Question 6.
In modern diving apparatus, a mixture of He and O₂ is used. Why? [AP ’16]
Answer:
Helium has very low solubility in blood. So it is used as a diluent for oxygen in modern diving apparatus.

Question 7.
Helium is heavier than hydrogen. Yet helium is used (instead of H₂) in filling balloons for meteorological observations. Why?
Answer:
Helium is non-inflammable and light gas. Hence it is used in filling balloons for mete-orological observations.

Question 8.
How is XeO₃ prepared ?
Answer:
XeO₃ is prepared by the hydrolysis of XeF₆.
XeF₆ + 3H₂O → XeO₃ + 6HF

Question 9.
Give the preparation of
a) XeOF₄
b) XeO₂F₂
Answer:
Partial hydrolysis of XeF₆ gives oxyfluorides.
a) XeF₆ + H₂O → XeOF₄ + 2HF
b) XeF₆ + 2H₂O → XeO₂F₂ + 4HF

Question 10.
Explain the structure of XeO₃. [TS ’16 ; IPE ’14]
Answer:
Xe undergoes sp³ hybridisation.

Unpaired electrons in d-orbital form dπ – pπ bonds with oxygen, sp³ orbitals form 3σ bonds with oxygen atoms. Because of the presence of lone pair, the molecule assumes pyramidal shape with a bond angle of 103°.

Question 11.
Noble gases are inert- Explain.
Answer:
The inertness to chemical reactivity is due to –

1. The noble gases except helium (1 s²) have completely filled ns²np⁶ electronic configuration in their valence shell.
2. They have high ionisation enthalpy and high positive electron gain enthalpy.

Question 12.
Write the name and formula of the first noble gas compound prepared by Bertlett.
Answer:
Xe⁺ [PtF₆]^–
Xenon hexa fluoro platinate

Question 13.
Explain the shape of XeF₄ on the basis of use VSEPR theory. [Mar. 2018 – AP]
Answer:
XeF₄ [Xenon tetrafluoride]:
shape is octahedral. But due to the presence of two lone pairs, the molecule assumes square planar structure.

Total number of electron pairs is 6. The Xe atom undergoes sp³d² Hybridisation. Hence shape is octahedral. But due to the presence of two lone pairs, the molecule assumes square planar structure.

Question 14.
Give the outer electronic configuration of noble gases.
Answer:
ns²np⁶

Question 15.
Why do noble gases form compounds with fluorine and oxygen only?
Answer:
Because fluorine and oxygen are highly electronegative, they can form compounds with noble gases.

Question 16.
How is XeOF₄ prepared? Describe its molecular shape.
Answer:
Partial hydrolysis of XeF₆ gives oxyfluorides
XeF₆ + H₂O → XeOF₄ + 2HF
XeOF₄ has square pyramidal.

Question 17.
What is the major source of helium?
Answer:
Natural gas

Question 18.
Which noble gas is radioactive? How is it formed?
Answer:
Radon. Radon is obtained as a decay product of Ra²²⁶.
\({ }_{88}^{226} \mathrm{Ra}\) → \({ }_{86}^{222} \mathrm{Rn}\) + \({ }_2^4 \mathrm{He}\)

Question 19.
Name the following.
a) most abundant noble gas in atmosphere
b) radioactive noble gas
c) noble gas with least boiling point
d) noble gas forming large number of compounds
e) noble gas not present in atmosphere.
Answer:
a) Argon
b) Radon
c) Helium
d) Xenon
e) Radon

Short Answer Questions (4 Marks)

Question 20.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained ? [AP ’17 ; IPE ’14] [Mar. 2018 – TS]
Answer:
Xenon fluorides are obtained by the direct combination of elements under appropriate conditions.

Question 21.
How are XeO₃ and XeOF₄ prepared?
Answer:
Hydrolysis of XeF₄ and XeF₆ gives XeO₃.
XeF₆ + 3H₂O → XeO₃ + 6HF
Partial hydrolysis of XeF₆ gives XeOF₄
XeF₆ + H₂O → XeOF₄ + 2HF

Question 22.
Give the formulae and describe the struc-tures of noble gas sp>ecies, isoelectronic with
a) ICl₄
b) IBr₂^–
c) BrO₃^–
Answer:
a) XeF₄
b) XeF₂
c) XeF₆

a) ICl₄ is isoelectronic with XeF₄. XeF₄ is square planar in which Xe undergoes sp³d² hybridisation. If has two lone pairs.

Structure of XeF₄: Square planar.

b) IBr₂^– is isoelectronic with XeF₂. XeF₂ is linear. Xe undergoes sp³d hybridisation. Three lone pairs occupy three vertices of triangle.

c) BrO₃^– is isoelectronic with XeF₆. In XeF₆, Xe undergoes sp³d³ hybridisation. XeF₆ has seven electron pairs (6 bonding pairs and one lone pair) XeF₆ thus has a distorted octahedral structure

Question 23.
Explain the reaction of the following with water. [AP ’15 ; IPE ’14]
a) XeF₂
b) XeF₄
c) XeF₆
Answer:
a) XeF₂ reacts with water and give Xe, HF and O₂
2XeF₂ + 2H₂O → 2Xe + 4HF + O₂

b) XeF₄ reacts with water and give Xe, XeO₃, HF and O₂
6XeF₄ + 12H₂O → 4 Xe + 2XeO₃ + 24 HF + 3O₂

c) XeF₆ reacts with water and give XeO₃, HF
XeF₆ + 3H₂O → XeO₃ + 6 HF

Question 24.
Explain the structures of [(AP ’17) (Mar. 2018 – T.S)]
a) XeF₂ and
b) XeF₄
Answer:
a) In XeF₂, Xe undergoes sp3d hybridisation.

There are three bond pairs and two lone pairs. The two bond pairs overlap end-end with 2p_z orbitals of two F atoms. Thus XeF₂ molecule has linear structure.

b) XeF₄ is square planar. Xe undergoes sp³d² hybridisation. [Mar. 2018 – A.P.]

There are two lone pairs and four bond pairs. The four bond pairs overlap end-end with 2p_z orbitals of four F atoms.

Question 25.
Explain the structure of [TS ’15]
a) XeF₆ and
b) XeOF₄.
Answer:
a) XeF₆ has distorted octohedral shapes. XeF₆ has seven electron pairs. (6 bonding and one lone pair). It has a distorted octahedral structure in the gas phase.

b) XeOF₄ has square pyramidal structure.

Xe in excited state undergoes sp³d² hybridisation. One hybrid orbital is occupied by a lone pair. Other five orbitals are occupied by unpaired electrons. Five orbitals form five σ bonds. The unpaired electrons in unhybridised’d’ orbital forms a π bond with oxygen. As the lone pair occupies axial position, XeOF₄ assumes square pyramidal structure.

Question 26.
Complete following:
a) XeF₂ + H₂O →
b) XeF₂ + PF₅ →
c) XeF₄ + SbF₅ →
d) XeF₆ + AsF₅ →
e) XeF₄ + O₂F₂ →
f) NaF + XeF₆ →
(Hint: NaF + XeF₆ → Na⁺[XeF₇]^–)
Answer:
a) XeF₂ is hydrolysed to give Xe, HF and O₂.
2XeF₂(s) + 2H₂O (l) → 2Xe (g) + 4HF (aq) + O₂(g)

b) Xenon fluorides react with fluoride ion acceptors to form cationic species.
XeF₂ + PF₅ → [XeF]⁺ [PF₆]^–

c) Xenon tetrafluoride reacts with fluoride ion to form cationic species.
XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]^–

d) XeF₆ + ASF₅ → [XeF₅]⁺ [AsF₆]^–

e) XeF₆ is prepared by the interaction of XeF₆ and O₂F₂ at 143 K.
XeF₄ + O₂F₂ → XeF₆ + O₂

f) XeF₆ reacts with fluoride ion donors to form fluoroanions.
XeF₄ + NaF → Na⁺ [XeF₇]^–

Question 27.
How are XeF₂ and XeF₄ prepared? Give their structures. [AP ’17]
Answer:
Xenon fluorides are obtained by the direct combination of elements under appropriate conditions.

Hydrolysis of XeF₄ and XeF₆ gives XeO₃.
XeF₆ + 3H₂O → XeO₃ + 6HF
Partial hydrolysis of XeF₆ gives XeOF₄
XeF₆ + H₂O → XeOF₄ + 2HF

Long Answer Questions (8 Marks)

Question 28.
How are XeF₂, XeF₂. Explain their reaction with water. Discuss their structures.
Answer:
XeF₂ is prepared by heating Xenon and Fluorine in a seated Nickel tube in the ratio 1 : 2 at 673K for 8hrs.

Reaction with water:
XeF₂ is hydrolysed to give Xe, HF and O₂
2XeF₂ (s) + 2H₂O (l) → 2Xe(g) + 4HF (aq) + O₂(g)

Structure:
Xe in the first excited state

Xe undergoes sp³d hybridisation. The sp³d orbitals assume trigonal bipyramidal geometry. Two fluorine atoms form 2π bonds with Xe.

The bond pairs occupy axial positions. XeF₄: XeF₄ is formed when Xe and F₂ mixed in the ratio 1 : 5, heated to 873K at 7 bar.

Reaction with water: XeF₄ hydrolysed with water to XeO₃.
6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂
Structure of XeF₄ : XeF₄ is square planar Xe in the second excited state

Xe undergoes sp³ d² hybridisation. Xe forms four o bonds with 4 fluorine atoms. Two lone pairs occupy axial positions. The four fluorine atoms occupy corners of a square.

Additional Questions – Answers

Question 1.
Why are the elements of group 18 known as noble gases?
Answer:
The group 18 elements have completely filled s and p orbitals in their valence shell. They react with a few elements only under certain conditions. Therefore they are known as noble gases.

Question 2.
Noble gases have very low boiling points. Why?
Answer:
Noble gases are monoatomic. There are no inter atomic forces except weak dispersion forces. Hence they are liquified at very low temperatures. Hence they have low boiling points.

Question 3.
Does the hydrolysis of XeF₆ lead to a redox reaction?
Answer:
No, the products of hydrolysis are XeF₄ and XeO₂F₂ where the oxidation states of all the elements remain the same.

Intext Questions – Answers

Question 1.
Why is Helium used in diving apparatus ?
Answer:
Helium acts as a diluent for oxygen. Hence it is used in diving apparatus.

Question 2.
Balance the following equation.
XeF₆ + H₆O > XeO₂ F₂ + HF
Answer:
XeF₆ + 2H₂O > XeO₂F₂ + 4HF

Question 3.
Why has it been difficult to study the chemistry of Radon?
Answer:
Radon is radioactive element with very short Half-life period. Hence its study is difficult.

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material Lesson 6(c) Group-17 Elements Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material Lesson 6(c) Group-17 Elements

Very Short Answer Questions (2 Marks)

Question 1.
Which halogen produces O₂ and O₃ on passing through water?
Answer:
Fluorine liberates O₂ and O₃ with water.
2F₂ + 2H₂O → 4HF + O₂
3F₂ + 3H₂O → 6HF + O₃

Question 2.
Interhalogen compounds are more reactive than the constituent halogens except fluorine- explain.
Answer:
The interhalogen compounds are more reactive than the halogens (except F₂). This is because the A-X bond in interhalogens is weaker than the X – X bond in the halogens.

Question 3.
What is the use of ClF₃?
Answer:

1. ClF₃ is a covalent liquid. It is used to make gaseous UF₆.
   3ClF₃ + U → UF₆ + 3ClF
2. It acts as powerful fluorinating agent for inorganic substances.
   4ClF₃ + 6MgO → 6MgF₂ + 2Cl₂ + 3O₂

Question 4.
Write two uses of ClO₂.
Answer:

1. ClO₂ is used as a bleaching agent for paper pulp and textiles.
2. Water treatment.

Question 5.
Why are halogens coloured ?
Answer:
The colour of halogens is due to absorption of visible light by their molecules resulting in excitation of outer electrons to higher energy levels. The excitation energy required by small fluorine atoms is large and that required by the large iodine atoms is least. By absorbing different quanta of radiation, they display different colours. F₂ is yellow, Cl₂ is greenish yellow, Br₂ is red and I₂ is violet.

Question 6.
Write the reactions of Cl₂ and F₂ with water. [IPE ’14]
Answer:

1. Chlorine dissolves in water to give hydrochloric acid and hypochlorous acid.
   Cl₂ + H₂O → HCl + HOCl
2. Fluorine oxidises water to oxygen and ozone.
   2F₂ + 2H₂O → 4HF + O₂
   3F₂ + 3H₂O → 6HF + O₃

Question 7.
With which neutral molecule, ClO^– is iso- electronic? Is that molecule a Lewis base ?
Answer:
ClF is isoelectronic with ClO^–. ClF is a Lewis base.

Question 8.
Arrange the following in the order of the property indicated for each set.
a) F₂,Cl₂, Br₂, I₂ – increasing bond dissociation enthalpy.
b) HF, HCl, HBr, HI – increasing acidic strength.
c) HF, HCl, HBr, HI – increasing boiling points.
Answer:
a) I₂ < f₂ < Br₂ < Cl₂
b) HF < HCl < HBr < HI
c) HCl< HBr < HI < HF

Question 9.
Electron gain enthalpy of fluorine is less than that of chlorine. Explain.
Answer:
Fluorine has lower electron gain enthalpy than chlorine. The lower value is attributed to small size of fluorine. Because of its small size, there is repulsion between the electron pairs already present and the electrons that are added. Hence, electron adding up tendency of F atom is less.
Therefore, Electron gain enthalpy of fluorine is – 333 kJ / mol. is less than that of chlorine (-349 kJ/mol).

Question 10.
HF is a liquid while HCl is gas -explain.
Answer:
HF is an associated liquid because of hydrogen bonds.
H – F ……………… H – F ………… H – F
HCl is not an associated liquid. Hence its boiling point is less than that of HF. Hence it is a gas.

Question 11.
Bond dissociation enthalpy of F₂ is less than that of Cl_2^– explain.
Answer:
Due to small size of fluorine atoms, the F – F distance is also small. Hence internuclear repulsion is high. The large repulsions bet-ween lone pair of electrons present on the two fluorine atoms weaken the bond. This repulsion is less in other molecules, because the atoms are large in size.

Question 12.
Write the formulae of the compounds, in which oxygen has positive oxidation states and mention the oxidation states of oxygen in them.
Answer:

Question 13.
What is the use of O₂F₂ and I₂O₅?
Answer:
O₂F₂ oxidises plutonium to PuF₆ and the reaction is used in removing plutonium as PuF₆ from spent nuclear fuel.
I₂O₅ is a good oxidising agent and is used in the estimation of carbon monoxide.

Question 14.
Write two uses of hydrogen chloride.
Answer:

1. HCl is used for ‘pickling’ metals that is removing oxide layers from the surface.
2. To make metal chlorides.
3. In the manufacture of dyestuffs.

Question 15.
Explain the reactions of Cl₂ with NaOH.
Answer:

1. With cold, dilute NaOH, chloride and hypochlorite are formed.
   
2. With hot, cone. NaOH, chloride and chlorate are formed.
   

Question 16.
What happens when Cl2 reacts with dry slaked lime ? [Mar. 2018 AP & TS ; AP 17]
Answer:
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O
Bleaching powder is formed when chlorine reacts with dry slaked lime.

Question 17.
Chlorine acts as an oxidising agent Explain with two examples. [AP ’16]
Answer:

1. Chlorine oxidises ferrous sulphate for ferric sulphate.
   2FeSO₄ + H₂SO₄ + Cl₂ → Fe₂ (SO₄)₃ + 2HCl
2. Chlorine oxidises SO₂ to SO₃ and in presence of water, H₂SO₄ is formed.
   SO₂ + 2H₂O + Cl₂ → H₂SO₄ + 2HCl

Question 18.
What is aqua-regia? Write its reaction with gold and platinum.
Answer:
When 3 parts cone. HC/ and 1 part cone. HN03 are mixed, aqua regia is formed.
\(\mathrm{Au}+4 \mathrm{H}^{+}+\mathrm{NO}_3^{-}+4 \mathrm{C} t \longrightarrow \mathrm{AuCl}_4^{-}+\mathrm{NO}+2 \mathrm{H}_2 \mathrm{O}\)
\(3 \mathrm{Pt}+16 \mathrm{H}^{+}+4 \mathrm{NO}_3^{-}+18 \mathrm{Cl} \longrightarrow 3 \mathrm{PtCl}_6^{2-}+4 \mathrm{NO}+8 \mathrm{H}_2 \mathrm{O}\)

Question 19.
How is Cl₂ manufactured by Deacon’s [AP & TS Mar. 19; (AP 17; TS 16)]
Answer:
Cl₂ is formed when HCl gas is oxidised by atmospheric oxygen in presence of CuCl₂ at 723K.

Question 20.
Chlorine acts as a bleaching agent only in the presence of moisture – Explain.
Answer:
Bleaching action of chlorine is due to oxidation. In presence of moisture, chlorine dis-solves in water to give nascent oxygen.
Cl₂ + H₂O → 2HCl + (O)
Coloured substance + (0) → Colourless substance

Question 21.
The decreasing order of acidic character among hypo halogen acids is HClO > HBrO > HIO. Give reason.
Answer:
The acidic character of HClO is due to its dissociation to give HCl.
HClO → HCl + (O)
Hypobromites and hypoiodides tend to disproportionate.
3BrO^– → 2Br^– + BrO₃^–
Hence stability of HBrO and HIO is less. Hence less acidic.

Question 22.
The acidic nature of oxoacids of chlorine is HOCl < HClO₂ < HClO₃ < HClO₄ – Explain. (Hint: HA + H₂O ⇌ H₃O⁺ + A^– conjugate base, greater the stability of A^–, lesser will be its basic strength or greater will be the tendency of HA to release H⁺. In other words, stronger will be the acid HA. Among the conjugate bases of oxoacids of chlorine, the stability order is OCl^– < ClO₂^– < ClO₃^– < ClO₄^–)
Answer:
Among the conjugate bases of oxoacids of chlorine, the stability order is
OCl^– < ClO₂^– < ClO₃^– < ClO₄^–
The more the stability of conjugate base, the stronger will be acid. Hence HClO₄ is a strong acid as ClO₄^– is stable. HOCl is weak as OCl^– is less stable.

Question 23.
What are interhalogen compounds? Give two examples.
Answer:
The binary compounds formed by the combination of halogens among themselves are called interhalogen compounds.
Ex : AX type ; ICl, Cl F
AX₃ type ; ClF₃, BrF₃
AX₅ type ; BrF₅, IF₅

Question 24.
Explain the structure of ClF₃.
Answer:
Excited chlorine

Chlorine undergoes sp³d hybridisation.
It assumes trigonal bipyramidal shape with two lone pairs of electrons. ClF₃ is planar, T- shaped with bond angle 90°. Lone pair distorts the angle slightly.

Question 25.
OF₂ should be called oxygen difluoride and not fluorine oxIde – Why?
(HInt : F Is more electronegative than Oxygen).
Answer:
As fluorine is more electronegative than oxygen, fluorine is in – 1 oxidation state. Hence the compounds of oxygen with fluorine are called fluorides. So, OF₂ is to be called oxygen difluoride.

Question 26.
Iodine is more soluble in KI than in water -Explain.
Answer:
Iodine combines with KI in aqueous solution to form KI₃.
KI + I₂ → KI₃
As complex formation increases solubility, solubility of I₂ is more in KI.

Question 27.
Among the hydrides of halogens (a) which is most stable? (b) which is most acidic? (c) which has lowest boiling point?
Answer:
a) HF is most stable,
b) HI is most acidic.
c) HCl has lowest boiling point.

Question 28.
Compare the bleaching action of Cl₂ and SO₂.
Answer:
Cl₂ acts as bleaching agent in presence of moisture. Cl₂ bleaches by oxidation.
Cl₂ + H₂O → HCl + (O)
Coloured substance + (0) → Colourless substance
SO₂ bleaches through reduction.
SO₂ + 2H₂O → H₂SO₄ + 2H
Coloured matter + 2[H] → Colourless matter

Question 29.
Give the oxidation states of halogens in the following. [TS ’15]
a) Cl₂O
b) ClO₂^–
c) KBrO₃
d) NaClO₄
Answer:
a) Let the oxidation state of Cl be x.
2x – 2 = 0
x = + 1
∴ O.S. of Cl in Cl₂O is + 1.

b) ClO₂^–; x – 4 = – 1
x = 4 – 1 = +3
∴ O.S. of Cl in ClO₂^– is + 3.

c) KBrO₃
O.S. of K = + 1; O.S. of O is -2
+1 + x – 6 = 0
x = + 5

d) NaClO₄
+1 + x – 8 = 0
x – 7 = 0
x = +7
O.S. of Cl in NaClO₄ is +7.

Question 30.
Describe the molecular shape of I₃^–. (Hint: Central iodine is of SP³d. – linear)
Answer:
Central Iodine atom is in sp³d hybridisation.

Short Answer Questions (4 Marks)

Question 31.
How can you prepare Cl₂ from HCl and HCl from Cl₂? Write the reactions.
Answer:

1. HCl is oxidised by MnO₂ to Cl₂.
   By heating manganese dioxide with concentrated HCl.
   MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
2. Chlorine reacts with H₂ to form HCl.
   H₂ + Cl₂ → 2HCl

Question 32.
Write balanced equations for the following. [IPE ’14]
a) NaCl is heated with cone. H₂SO₄ in the presence of MnO₂
b) Chlorine is passed into a solution of NaI in water.
Answer:
a) Cl₂ gas liberated.
4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂
b) I₂ is liberated.
2NaI + Cl₂ → 2NaCl + I₂

Question 33.
Explain the structures of
a) BrF₅ and
b) IF₇.
Answer:
a) BrF₅: Central Bromine atom in BrF₅ undergoes sp³d² hybridisation. It has distorted octahedral shape.

b) IF₇ : Central Iodine atom in IF₇ undergoes sp³d³ hybridisation. It has pentagonal bipyramidal structure.

Question 34.
Write a short note on hydrogen halides.
Answer:
Halogens react with hydrogen forming hydrogen halides.
H₂ + X₂ → 2HX
Properties:

1. Stability decreases from HF to HI.
   HF > HCl > HBr > HI
2. The acidic strength increases in the order HF < HC/ < HBr < HI.
3. The boiling points HCl < HBr < HI < HF
   HF is associated liquid due to hydrogen bonding
   H – F …………. H – F …………. H – F

Uses:

1. HF is used in etching of glass.
2. HCl is used in the manufacture of chlorine, NH₄Cl and glucose.
3. HCl is used for extracting glue from bones and purifying bone black.

Question 35.
How is chlorine obtained in the laboratory ? How does it react with the following ? [(Mar. 18 AP) (AC & TS 16)]
a) cold dil. NaOH
b) excess NH₃
c) KI
Answer:
By heating concentrated hydrochloric acid with Manganese dioxide.
MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
Reactions:
a) When chlorine is passed through cold dil. NaOH, chloride and hypochlorite are formed.
2NaOH + Cl₂ → NaCl + NaOCl + H₂O

b) Chlorine gives nitrogen and ammonium . chloride with excess ammonia.
8NH₃ + 3Cl₂ → 6NH₄Cl + N₂

c) When Cl₂ gas is passed through KI, I₂ is liberated.
2KI + Cl₂ → 2KCl + I₂

Question 36 .
What are interhalogen compounds? Give some examples to illustrate the definition. How are they classified?
Answer:
When two different halogens react with each other, interhalogen compounds are formed.
They are classified as AX, AX₃, AX₅ and AX₇ where A is halogen of larger size and X is smaller size.
They are prepared by direct combination or by the action of halogen on lower inter-halogen compounds.

Example:
AX Type : ClF, BrF, ICl
AX₃Type : ClF₃, BrF₃, ICl₃
AX₅ Type : BrF₅, IF₅
AX₇ Type: IF₇

Long Answer Questions (8 Marks)

Question 37.
How is ClF₃ prepared? How does it react with water? Explain its structure.
Answer:
ClF₃ is prepared by the reaction of Cl₂ with excess of Fluorine.

ClF₃ is a colourless gas. It has bent T-shape structure.

Reaction with water :
Chloric acid is formed when ClF₃ undergoes hydrolysis.
ClF₃ + 2H₂O → 3HF + HClO₂
Structure:

Excited state of chlorine
Chlorine undergoes sp³d hybridisation. There are two lone pairs and three bond pairs.

ClF₃ is T – shaped. The bond pairs and lone pairs occupy corners of trigonal bipyramid. The two lone pairs occupy the equatorial positions to minimise the repulsions.

Question 38.
How is chlorine prepared in the laboratory ? How does it react with the following ? [TS ’15]
a) Iron
b) hot cone. NaOH
c) acidified FeSO₄
d) Iodine
e) H₂S
f) Na₂S₂O₃
Answer:
Chlorine is prepared in the laboratory by heating concentrated Hydrochloric acid with Manganese dioxide.
MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
It can also be prepared by heating a mixture of sodium chloride and concentrated H₂SO₄.
4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂
Reactions:
a) It oxidises Fe to Fe3+.
2Fe + 3Cl₂ → 2FeCl₃

b) When Cl₂ is passed through hot cone. NaOH sodium chlorate is formed.
6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O

c) It oxidises acidified ferrous sulphate to ferric sulphate.
2FeSO₄ + H₂SO₄ + Cl₂ → Fe₂(SO₄)₃ + 2HCl

d) With I₂ in presence of water, it forms iodic acid.
I₂ + 6H₂O + 5Cl₂ → 2HIO₃+ 10HCl

e) It reacts with H₂S to form HCl.
H₂S + Cl₂ → 2HCl + S ↓

f) Hypo reacts with Cl₂ to form HCl and precipitates sulphur.
Na₂S₂O₃ + Cl₂ + H₂O → Na₂SO₄ + 2HCl + S ↓

Question 39.
Discuss anomalous behaviour of fluorine.
Answer:
Fluorine, the first element of halogen group shows anomalous properties.

1. Its high reactivity is due to small bond dissociation enthalpy of F – F bond and high electronegativity.
2. The electronegativity, ionisation enthalpy and electrode potentials are higher them expected.
3. Bond dissociation and electron gain enthalpy are quite lower than expected.
4. The anomalous behaviour of fluorine is due to its small size, highest electronega-tivity, low F- F bond dissociation enthalpy and non-availability of d-oribtals in the valence shell.
5. HF is a liquid while other hydrogen halides are gases.
6. Most of the reactions of fluorine are exothermic.
7. AgCl is insoluble in water whereas AgF is soluble.
8. CaCl₂ is soluble whereas CaF₂ insoluble.

Question 40.
How is chlorine prepared by electrolytic method? How does it react with a) NaOH and b) NH₃ under different conditions. [AP ’16, ’15]
Answer:
Chlorine is obtained by the electrolysis of brine solution. Chlorine is liberated at anode.

Brine solution is taken in a perforated U-shaped steel vessel. Graphite rod acts as anode, steel vessel acts as cathode. H₂ and NaOH are by products.

NaCl → Na⁺ + Cl^–
At Anode : 2Cl^– – 2e → Cl₂
2H₂O + 2e → 2OH^– + H₂
Na⁺ + OH^– → NaOH
Reactions:

1. When chlorine gas is passed through cold, dil. NaOH, NaCl and NaOCl are formed.
   2NaOH + Cl₂ → NaCl + NaOCl + H₂O
2. When Cl₂ gas is passed through hot concentrated NaOH, sodium chloride and sodium chlorate are formed.
   6NaOH + 3Cl₂ → 5NaCl + NaClO₃+ 3H₂O

b) With excess ammonia, Nitrogen gas is evolved.
8NH₃ + 3Cl₂ → 6NH₄Cl + N₂
When chlorine is in excess, NCl₃ and HCl are formed.
NH₃ + 3Cl₂ → NCl₃ + 3HCl

Question 41.
Write the names and formulae of oxoacids of chlorine. Explain their structures and acidic relative nature.
Answer:
Hypochlorous acid HOCl
Chlorous acid HClO₂
Chloric acid HClO₃
Perchloric acid HClO₄
Acidic character : Among the different oxoacids of chlorine, the acidic character follows the order.
HOCl < HClO₂ < HClO₃ < HClO₄
When these acids, dissolve in water their conjugate bases OCl^–, ClO₂^–, ClO₃^– and ClO₄^– are formed. As the oxidation of chlorine increases, number of n bonds increase and delocalisation increases. As delocalisation of n electrons increases, stability of conjugate base increases. Hence the ability of anion to accept proton decreases. Hence the basic character of anions increases in the order.
ClO₄^– < ClO₃^–– < ClO₂^– < ClO^–
ClO^– is strong base, its conjugate acid HOCl is weak.
ClO₄^– is weak base, its conjugate acid HClO₄ is strong.
Hence the acidic character increases and the order is
HOCl < HClO₂ < HClO₃ < HClO₄.

Structures: In all these acids, chlorine uses sp³ hybrid orbitals.
1) HOCl: It has linear shape no dπ – pπ bonds with three line pair of electrons.

2) HClO₂ : The anion of the acid is ClO₂^–. It has a V shape. There are two lone pairs in chlorine. The electron in 3d orbitals forms a π bond with oxygen.

HClO₃ : ClO₃^– is pyramidal. Two unpaired electrons in 3d form two dπ – pπ bonds with oxygen. There is one lone pair.

HClO₄ : Perchlorate ion has tetrahedral shape. Three dπ – pπ bonds are formed.

As there is more delocalisation the stability of ions increases in the order.
ClO^–< ClO₂^– < ClO₃^– < ClO₄^–

Intext Questions – Answers

Question 1.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and cl₂.
Answer:
Because of low enthalpy of dissociation of F – F bond and high hydration enthalpy, F₂ acts as strong oxidising agent than chlorine.

Question 2.
Give two examples to show the anoma¬lous behaviour of Flurine.
Answer:
HF is a liquid while others (HCl, HBr, Hr) are 4. gases. AgCl is insoluble in water whereas AgF is soluble.

Question 3.
Sea is the greatest source of some Halogens. Comment.
Answer:
Sea water contains chlorides, bromides and Iodides of sodium, potassium, magnesium and calcium. The deposits of dried up seas contain sodium chloride and carnalite. Certain forms of marine life and various seaweeds contain Iodine. Hence we can say that sea is the greatest source of some Halogens.

Question 4.
Give the reason for the bleaching action of Cl₂.
Answer:
Chlorine bleaches by oxidation. Coloured substance + (0) → Colourless substance

Question 5.
Name two poisonous gases which can be prepared from chlorine gas.
Answer:
Phosgene, Tear gas, Mustard gas.

Question 6.
Why is ICl more reactive than I₂?
The ICl bond is weaker than I – I bond.

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 2nd Lesson Solutions Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 2nd Lesson Solutions

Very Short Answer Questions (2 Marks)

Question 1.
Define the term solution.
Answer:
Homogeneous mixture of two or more than two components is called solution.

Question 2.
Define molarity.
Answer:
Molarity (M) is defined as number of moles of solute dissolved in one litre of (or one cubic decimetre) solution.

Question 3.
Define molality.
Answer:
Molality (m) is defined as the number of moles of the solute present in one kilogram (kg) of the solvent.

Question 4.
Give an example of a solid solution in which the solute is solid.
Answer:
In solid solution solute is solid. Here sol-vent is also solid. e.g., copper dissolved in gold (alloys).

Question 5.
Define mole fraction. (Mar. 2018 TS)
Answer:
Mole fraction is the ratio of number of moles of one component to the total number of moles of all the components in the solution.
Mole fraction of a component

Question 6.
Define mass percentage solution.
Answer:
The mass percentage of a component of a solution is defined as
Mass % of a component
Mass % of the component

Question 7.
What is ppm of a solution ?
Answer:
Parts per million
Number of parts of the

This method is convenient when a solute is present in trace amounts.

Question 8.
What role do the molecular interactions play in a solution of alcohol and water ?
Answer:
The interaction between alcohol and alcohol is hydrogen bonding and H₂O and H₂O is also hydrogen bonding. After mixing the interaction between alcohol and H₂O is also hydrogen bonding but less than the hydrogen bonding in alcohol and H₂O separately. So after mixing the interaction between alcohol and water decreases and the vapour pressure of solution increases forming low boiling point azeotrope.

Question 9.
State Raoult’s law. (AP & TS 16) (IPE 14) (Mar. 2018 . AP & TS)
Answer:
For any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.

Question 10.
State Henry’s law.
Answer:
The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution.
(or)
Mole fraction of gas in the solution is proportional to the partial pressure of the gas over the solution.
(or)
The partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.
P = K_H × x
K_H is the Henry’s law constant.

Question 11.
What is Ebullioscopic constant?
Answer:
The elevation in the boiling point of one molal solution i.e., when one mole of solute is dissolved in 1 kg of solvent, is called boiling point elevation constant or molal elevation constant or ebullioscopic constant.
ΔT_b = K_b m.
unit of K_b = K kg mol^-1

Question 12.
What is Cryoscopic constant?
Answer:
The depression in freezing point of one molal solution i.e., when one mole of solute is dissolved in 1 kg of solvent, is called freezing point depression constant or molal depression constant or cryoscopic constant.
ΔT_f = K_f m.
unit of K_f = K kg mol^-1

Question 13.
Define Osmotic pressure. (AP 16, 15) (Mar. 2018 – AP)
Answer:
When a solution is separated from a solvent by a semipermeable membrane or if a dilute solution is separated from concentrated solution by a semipermeable membrane, the pressure that just prevents passage of solvent into solution or solvent from dilute solution into concentrated solution is called osmotic pressure.
(or)
The extra pressure that is to be applied on the solution side when the solution and solvent are separated by a semipermeable membrane to stop osmosis.

Question 14.
What are isotonic solutions ? TS Mar. 19; (AP 17, 15; TS 16)
Answer:
Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

Question 15.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water,
(i) Phenol
(ii) Toluene
(iii) Formic acid
(iv) Ethylene glycol
(v) Chloroform
(vi) Pentanol.
Answer:
Insoluble : Chloroform, toluene.
Partially soluble : Phenol, pentanol.
Highly soluble: Formic acid, ethylene glycol.

Question 16.
Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 gm of C₉H₈O₄ is dissolved in 450g of CH₃CN.
Answer:
Mass percentage of aspirin.

Question 17.
Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250ml of 0.15M solution in methanol.
Answer:
Molecular mass of benzoic acid (C₆H₅COOH) = 12 × 6 + 5 + 12 + 32 + 1 = 122 g mol^-1
moles of benzoic = M × V
= 250 × 0.15 = 37.5
= 37.5 × 10^-3 mol
Amount of benzoic acid = Moles × MW₂
= 37.5 × 10^-3 × 122
= 4.575 g.

Question 18.
The depression in freezing point of water observed for the same amount of acetic acid, dichloro – acetic acid and trichloro acetic acid increases in the order given above. Explain briefly.
Answer:
Increasing order of ΔT_f (depression in freezing point) is Acetic acid < dichloroacetic acid < trichloroacetic acid.
Due to more electronegativity of Cl atom it exerts -1 (inductive effect) consequently dichloroacetic acid is the stronger acid than acetic acid. As the number of chlorine atoms increases the inductive effect also increases. Hence trichloro acetic acid is stronger acid than dichloroacetic acid.

Question 19.
What is van’t Hoffs factor ‘i’ and how is it related to ‘α’ in the case of a binary ele-ctrolyte (1 : 1) ?
Answer:
van’t Hoff introduced a factor ‘i’ known as the van’t Hoff factor to account the extent of dissociation or association. The factor ‘i’ is defined as

If α represents the degree of association of the solute for an equilibrium.
nA \(\rightleftharpoons\) An
Total number of particles at equilibrium is 1 – α + \(\frac{\alpha}{2}\) = 1 – \(\frac{\alpha}{2}\)
Then i = \(\frac{\alpha / 2}{1 \cdot 1}\) = \(\frac{\alpha}{2}\)
For dissociation

Total moles of particles aren(1 – α + α + α)
= n(1 + α)
i = \(\frac{n(1+\alpha)}{n}\) = 1 + α

Question 20.
What is relative lowering of vapour pressure ? (AP Mar. 19)
Answer:
The ratio of the lowering of vapour pressure of a solution containing a non – volatile solute to the vapour pressure of pure solvent is called relative lowering of vapour pressure.
It can be shown as R.L.V.P. = \(\frac{p^{\circ}-p}{p^{\circ}}\)
Here p° is the vapour pressure of pure solvent, p is the vapour pressure of solution containing non – volatile solute.

Question 21.
Calculate the mole fraction of H₂SO₄ in a solution containing 98% H₂SO₄ by mass. (Mar. 2018 – TS)(IPE 14)
Answer:
Mass of water = 2 gm.
Moles of water = \(\frac{2}{18}\) = \(\frac{1}{9}\) (M.Wt. of H₂O = 18)
Mass of H₂SO₄ = 98 gm.
Moles of H₂SO₄ = \(\frac{98}{98}\)
= 1 (Mol. Wt. of H₂SO₄ = 98)
Moles fraction of H₂SO₄ = \(\frac{1}{1+\frac{1}{9}}\) = \(\frac{9}{10}\) = 0.9

Question 22.
How many types of solutions are formed ? Give an example for each type of solution.
Answer:
Depending on the type of solvent three types of solutions will be formed.

Question 23.
Define mass percentage, volume percentage and mass to volume percentage solutions.
Answer:
Mass percentage (w/w) : The mass percentage of a component of a solution is defined as
Mass % of a component
Mass of the component

Volume percentage (V/V) : The volume percentage is defined as
Volume % of a component

Mass to volume percentage (w/v): It is the mass of solute dissolved in 100 ml of the solution.

Question 24.
Concentrated nitric acid used in the laboratory work is 68% nitric acid by in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL^-1?
Answer:
68% by mass implies that 68 g of HNO₃ is present in 100 g of solution.

Question 25.
A solution of glucose in water is labelled as 10% w/w. What would be the molarity of the solution?
Answer:
Wt. of glucose = 1o g
Wt. of solution = 100 gm
Wt. of water = 100 – 10 = 90 gm.
Considering density of water as 1 gm mL^-1
Volume of water = 90 mL

Question 26.
A solution of sucrose In water is labelled as 20% w/w. What would be the mole fraction of each component in the solution?
Answer:
Wt. of sucrose = 20
Moles of sucrose = \(\frac{20}{342}\) = 0.0585
Wt. of solution = 100 g
Wt. of water = 100 – 20 = 80 g
Moles of water = \(\frac{80}{18}\) = 4.44
Mole fraction of sucrose = \(\frac{0.0585}{0.0585+4.44}\)
= 0.013
Mole fraction of water = \(\frac{4.44}{0.0585+4.44}\)
= 0.987

Question 27.
How many ml of 0.1 MHCl is required to react completely with 1.0 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both ?
Answer:
Let Na₂CO₃ be x g

Since, given mixture contains equimolar amount.
∴ \(\frac{\mathrm{x}}{106}\) = \(\frac{1-x}{84}\) (or) x = 0.56
Total HCl required
= \(\left[\frac{2 x}{106}+\frac{1-x}{84}\right]\) = \(\frac{2 \times 0.56}{106}\) + \(\frac{1-0.56}{84}\)
= 0.01578 mol
If V is the volume of HCl required, then
V_(L) × 0.1 = 0.01578
(or) V_(L) = 0.1578L (or) 157.8 mL.

Question 28.
A solution is obtained by mixing 300g of 25% solution and 400g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Answer:
Mass of solute in 300 g of 25% solution.
= \(\frac{300 \times 25}{100}\) = 75 g
Mass of solute in 400 g of 40% solution
= \(\frac{400 \times 40}{100}\) = 160 g
Total mass of solute = 75 + 160 = 235 g
Total mass of solution = 700 g
% solute in final solution = \(\frac{235 \times 100}{700}\)
= 33.5%
% of water in final solution = 100 – 33.5
= 66.5%

Question 29.
An antifreeze solution is prepared from 222.6g of ethylene glycol (C₆H₆O₂) and 200g of water (solvent). Calculate the molality of the solution.
Answer:
Mass of glycol (w₁) = 222.6 g
Moles of glycol (n₁)

=
= \(\frac{222.6}{62}\)
= 3.59 mol
Mass of water (w₂) = 200 g = \(\frac{200 \times 1 \mathrm{~kg}}{1000}\)
= 0.2 kg
Mass of solution (w₁ + w₂) = 200 + 222.6
= 422.6 g
Density of solution (d) = 1.072 g mL^-1
Volume of solution (V)
=
= \(\frac{422.6}{1.072}\)
= 394.22 mL (or) 0.3942 L
Molality (m) = \(\frac{\mathrm{n}_1}{\mathrm{w}_2 \mathrm{~kg}}\) = \(\frac{3.59 \mathrm{~mol}}{0.2 \mathrm{~kg}}\)
= 17.95 mol kg
Molarity (M) = \(\frac{n_1}{V \text { in lit }}\) = \(\frac{3.59}{0.3942}\) = 9.1 mol L^-1

Question 30.
Why do gases always tend to be less soluble in liquids as the temperature is raised ?
Answer:
Dissolution of gases is exothermic process. It is because dissolution of a gas in a liquid decreases the entropy (ΔS < 0). Thus increase of temperature tends to push the equilibrium (Gas + solvent \(\rightleftharpoons\) solution, ΔH = -Ve) in the backward direction, thereby, suppressing the dissolution.

Question 31.
What is meant by positive deviations from Raoult’s law and how is the sign of Δ_mix H related to positive deviation from Raoult’s law ?
Answer:
The solutions which do not obey Raoult’s law and are accompanied by change in enthalpy and change in volume during their formation are called non – ideal solutions.

In the solutions showing positive deviations the partial vapour pressure of each component (say A and B) of solution is greater them the vapour pressure as expected according to Raoult’s law. In this type of solutions the solvent – solvent and solute – solute interactions are stronger than solvent – solute interactions since in the solution, the interactions among molecules become weaker, their escaping tendency increases which results in the increase in their partial vapour pressures. In such solutions total vapour pressure of the solution is also greater than the vapour pressure required according to the Raoult’s law.

For this type of non – ideal solutions exhibiting positive deviations.

1. P_A < \(P_A^0 x_A\); P_B > \(\mathrm{P}_{\mathrm{B}}^0 \mathrm{x}_{\mathrm{B}}\)
2. Δ_mixH = +ve
3. Δ_mixV = +ve

Question 32.
What is meant by negative deviation from Raoult’s law and how is the sign of Δ_mixH related to negative deviation from Raoult’s law?
Answer:
The solutions which do not obey Raoult’s law and are accompanied by change in enthalpy and change in volume during their formation are called non – Ideal solutions. In the solutions showing negative deviations the partial vapour pressure of each component of solution is less than the vapour pressure as expected according to Raoult’s law. In this type of solutions solvent – solvent and solute – solute interactiòns are weaker than that of solvent – solute interactions.

So the interactions among molecules in solution become stronger. Hence the escaping tendency of molecules decrease which results in the decrease in their partial vapour pressure. In such solutions total vapour pressure of the solution is also less than the vapour pressure expected according to Raoult’s law.

For these solutions exhibiting negative deviations

1. 
2. Δ_mix H = -ve
3. Δ_mixV = -ve

Question 33.
The vapour pressure of water is 12.3 k P_a at 300K. Calculate the vapour pressure of 1 molar solution of a non – volatile solute in it.
Answer:
Vapour pressure of water \(\mathrm{P}_{\mathrm{H}_2}^0 \mathrm{O}\) 12.3 k Pa
In 1 molar solution
Moles of water n_H2O = \(\frac{1000}{18}\) = 55.5 mol
Moles of solute n_B = 1 mol
Mole fraction of H₂O (X_H2O) = \(\frac{\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}}{\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}+\mathrm{n}_{\mathrm{B}}}\)
= \(\frac{55.5}{55.5+1}\) = 0.982

Question 34.
Calculate the mass of a non – volatile solute (molar mass 40g mol^-1) which should be dissolved in 114g Octane to reduce its vapour pressure to 80%. (TS 16; IF’ 14)
Answer:
Vapour pressure of solution (p) = 80% of \(\mathrm{P}_{\mathrm{A}}^0\) = 0.8\(\mathrm{P}_{\mathrm{A}}^0\)
Let the mass of solute be w g
∴ Moles of solute n_B = \(\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}}\) = \(\frac{W}{40}\) mol

Question 35.
A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water If freezing point of water is 273.15K.
Answer:
Molarity of sugar solution (m) = \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}\)
(or) m = \(\frac{5 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{342 \mathrm{~g} \mathrm{~mol}^{-1} \times 95 \mathrm{~g}}\) = 0.154 mol kg^-1
ΔT_f = 273.15 – 271 = 2.15°
Now ΔT_f = K_f × m or K_f = \(\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{m}}\) = \(\frac{2.15}{0.154}\) = 13.96
Molailty of glucose solution (m)
= \(\frac{5 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{180 \mathrm{~g} \mathrm{~mol}^{-1} \times 95 \mathrm{~g}}\) = 0.292 mol kg^-1
ΔT_f = K_fm= 13.96 × 0.292 = 4.08°
Freezing point of glucose solution
= 273.15 – 4.08 = 269.07 K.

Question 36.
If the osmotic pressure of glucose solution is 1.52 bar at 300K. What would be its concentration if R = 0.083L bar mol^-1 K^-1?
Answer:
Temperature T = 300 K
Osmotic pressure = 1.52 bar
R = 0.083 L bar mol^-1 K^-1
π = CRT (or) C = \(\frac{\pi}{\mathrm{RT}}\)
= \(\frac{1.52 \mathrm{bar}}{0.083 \mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1} \times 300}\)
∴ = 0.061 mol L^-1

Question 37.
Vapour pressure of water at 293K is 17.535mm Hg. Calculate the vapour pressure of the solution at 293K when 25g of glucose is dissolved in 450 g of water. (AP Mar. 19)
Answer:
Vapour pressure of water \(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}^{\mathrm{O}}\) = 17.535 mm.
Let vapour pressure of solution be p_s

Question 38.
How is molar mass related to the elevation in boiling point of a solution ?
Answer:
To calculate the molar mass of an unknown non – volatile compound a known mass (say W_Bg) of it is dissolved in a known mass (say W_A g) of some suitable solvent and elevation in its boiling point (ΔT_b) is determined. Let M_B be the molar mass of the compound. Then
Molarity of solution m = \(\frac{W_B}{M_B} \times \frac{1000}{W_A}\)
We know, ΔT_b = K_b × m = K_b × \(\frac{W_B}{W_A}\) × \(\frac{1000}{M_B}\)
M_B = \(\frac{\mathrm{K}_{\mathrm{b}} \times \mathrm{W}_{\mathrm{b}} \times 1000}{\mathrm{~W}_{\mathrm{A}} \times \Delta \mathrm{T}_{\mathrm{b}}}\)
Knowing K_b, W_B, W_A and the molar mass of the compound can be calculated
from the above relation. This method is known as ebullioscopic method.

Question 39.
What is an Ideal solution?
Answer:
An ideal solution may be defined as the solution which obeys Raoults law over the entire range of concentration and temperature.

According to Raoult’s law, the vapour pressure of a volatile component (p_A) of the solutions is equal to the product of its mole fraction (x_A) in solution and vapour pressure in pure state \(\mathrm{P}_{\mathrm{A}}^0\).
P_A = \(\mathrm{p}_{\mathrm{A}}^0 \mathrm{x}_{\mathrm{A}}\)

The formation of ideal solution neither involve any change in enthalpy nor in volume. An ideal solution,

1. should obey Raoults law i.e., p_A = \(p_A^0 x_A\) and p_B = \(p_B^0 x_B\)
2. Δ_mixH = 0
3. Δ_mixV = 0

In ideal solutions the solvent – solvent and solute – solute interactions are almost the same type as solvent – solute interactions.

Question 40.
What is relative lowering of vapour pressure ? How is it useful to determine the molar mass of a solute ?
Answer:
When a non – volatile solute such as urea, glucose etc., is dissolved in a volatile solvent such as water, the vapour pressure of solution will be less than that of pure solvent. This is known as lowering of vapour pressure.

The ratio of lowering of vapour pressure to that vapour pressure of pure solvent is known as relative lowering of vapour pressure.
\(\frac{\Delta \mathrm{p}}{\mathrm{p}_{\mathrm{A}}^0}\) = Relative lowering of vapour pressure
Δp = \(\mathrm{p}_{\mathrm{A}}^0-\mathrm{p}_{\mathrm{A}}\) = lowering of vapour pressure
\(\mathrm{p}_{\mathrm{A}}^0\) = vapour pressure of pure solvent

Determination of molar mass of a solute:
According to Raoult’s law the relative lowering of vapour pressure i equal to the mole fraction of the solute.
\(\frac{\Delta \mathrm{p}}{\mathrm{p}_{\mathrm{A}}^0}\) = x_B
x_B = mole fraction of solute,

If a known mass (W_B) of the solute is dissolved in a known mass (W_A) of solvent to prepare a dilute solution and the relative lowering of vapour pressure Is determined experimentally the molar mass of solvent (M_A) is known, the molar mass of solute M_B can be determined as follows.

In this equation all the parameters except M_B are known and hence M_B can be calculated.

Question 41.
How is molar mass related to the depre-ssion in freezing point of a solution ?
Answer:
To determine the molar mass of an unknown non – volatile compound a known mass (say W_Bg) of it is dissolved in a known mass (say W_Ag) of some suitable solvent and depression in the freezing point (ΔT_f) is determined. Let M_B be the molar mass of the compound. Then
Molarity of the solution m = \(\frac{W_B}{M_B} \times \frac{1000}{W_A}\)
We know ΔT_f = K_f × m = K_f × \(\frac{W_B}{W_A}\) × \(\frac{1000}{M_B}\)
M_B = \(\frac{\mathrm{K}_{\mathrm{b}} \times \mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{~W}_{\mathrm{A}} \times \Delta \mathrm{T}_{\mathrm{f}}}\)
Knowing K_f, W_B, W_A and ΔT_f, the molar mass of the compound can be calculated from the above relation. This method is called cryoscopic method.

Long Answer Questions (8 Marks)

Question 42.
An aqueous solution of 2% non – volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molecular mass of the solute ?
Answer:
The vapour pressure of pure water
\(\mathbf{p}_{\mathrm{A}}^0\) = 1 atm = 1.013 bar
Vapour pressure of solution (p) = 1.004 bar
W_B = 2g = W_A + W_B = 100g; W_A = 98g
Now \(\frac{\mathbf{p}_{\mathrm{A}}^0-\mathrm{P}}{\mathbf{p}_{\mathrm{A}}^0}\) = \(\frac{\mathrm{n}_{\mathrm{B}}}{\mathrm{n}_{\mathrm{A}}}\) = \(\frac{W_B / M_B}{W_A / M_A}\)
or M_B = \(\frac{W_B M_A}{W_A\left(\Delta P / P_A^0\right)}\) = \(\frac{2 \times 18 \times 1.013}{98 \times(1.013-1.004)}\)
= 41.35 g. mol^-1

Question 43.
Heptane and Octane form an ideal solution. At 373K the vapour pressure of the two liquid components are 105.2 kP_a and 46.8 kP_a respectively. What will be the vapour pressure of a mixture of 26.0 g heptane and 35g of octane ?
Answer:
Moles of C₇H₁₆(n_H) = \(\frac{W_{(H)}}{M_{(H)}}\) = \(\frac{25}{100}\) = 0.25 mol
Moles of C₈H₁₈ (n_O) = \(\frac{W_{(0)}}{M_{(0)}}\) = \(\frac{35}{114}\) = 0.307 mol
Mole fraction of C₇H₁₆ = (X_H) = \(\frac{0.25}{0.25+0.307}\)
= \(\frac{0.25}{0.557}\) = 0.466
Mole fraction of C₈H₁₈ (X₀) = 1 – 0.449
= 0.534
Vapour pressure of heptane (pH)
\(\mathrm{p}_{\mathrm{H}}^{\mathrm{O}}\) × X_H
= 105.2 \(\mathrm{Kp}_{\mathrm{a}}\) × 0.449
= 47.2348 \(\mathrm{Kp}_{\mathrm{a}}\) ≅ 49.02 \(\mathrm{kp}_{\mathrm{a}}\)
Vapour pressure of octane (P_O)
= \(P_{\mathrm{H}}^O\) × X_O
= 46.8 \(\mathrm{kp}_{\mathrm{a}}\) × 0.551 = 24.99 \(\mathrm{kp}_{\mathrm{a}}\)
Total vapour pressure P_Total = P_H + P_O
= 49.02 + 24.99 = 74.09 \(\mathrm{kp}_{\mathrm{a}}\)

Question 44.
A solution containing 30g of non – volatile solute exactly in 90g of water has a vapour pressure of 2.8 kPa at 298k. Further 18g of water is then added to the solution and the new vapour pressure becomes 2.9 \(\mathrm{kp}_{\mathrm{a}}\) at 298K. Calculate
(i) The molar mass of the solute and
(ii) Vapour pressure of water at 298k.
Answer:
Moles of solute (n_B) = \(\frac{30}{M_B}\)
Moles of H₂O(nH₂O) = \(\frac{90}{18}\) = 5 mol
Mole fraction of H₂O

After adding 18g(= 1 mol) of water to solution new mole fraction of water(\(\mathrm{X}_{\mathrm{H}_2 \mathrm{O}}^{\prime}\)) is

Question 45.
Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20g of Benzene (C₆H₆), 1g of AB₂ lowers the freezing point by 2.3K whereas 1.0g of AB₄ lowers it by 1.3K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate atomic masses of A and B.
Answer:
We know

Let the atomic weight of A = x
Let the atomic weight of B = y
x + 2y = 110.86
x + 4y = 196.15
Solving for x and y, we get
y = 42.64, x = 25.58
∴ Atomic weight of A = 25.8 u
Atomic weight of B = 42.64 u

Question 46.
Calculate the depression In the freezing point of water when 10g of CH₃CH₂CHClCOOH is added to 250g of water. K_a = 1.4 × 10^-3, K_f = 1.86K kg mol^-1.
Answer:
Calculation of molarity of the solution:
Mass of the solution = 250 + 10 = 260 g

Calculation of van’t Hoffs factor (i) : Let degree of dissociation of acid be α since acid is monobasic acid, therefore α and K_a of acid are related as
α = \(\sqrt{K_{\mathrm{a}} / \mathrm{c}}\) = \(\sqrt{\frac{1.4 \times 10^{-3}}{0.284}}\) = 0.07
Van’t Hoff factor and degree of dissociation are related as
α = \(\frac{i-1}{m-1}\) = \(\frac{i-1}{2-1}\) or i = 1 + α = 1 + 0.07
(or) i = 0.7
Calculation of depression in freezing point ΔT_f
ΔT_f = i × K_f × m = \(\frac{1.07 \times 1.86 \times 10 \times 1000}{122.5 \times 250}\)
= 0.649 = 0.65°C

Question 47.
19.5 g of CH₂FCOOH is dissolved in 500g of water. The depression in freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Answer:
Mass of solution = 500 + 19.5 = 519.5 g.

Moles of CH₂F COOH (n_B) = \(\frac{W_B}{M_B}\) = \(\frac{19.5}{78}\)
= 0.25 mol
Molarity of solution (M) = \(\frac{n_B}{V_{(L)}}\) = \(\frac{0.25}{0.462}\)
= 0.541 moL^-1
Mass of water (W_A) = 500 g = 0.5 kg
Molarity of solution (m) = \(\frac{\mathrm{n}_{\mathrm{B}}}{\mathrm{W}_{\mathrm{A}} \mathrm{kg}}\) = \(\frac{0.25 \mathrm{~mol}}{0.25 \mathrm{~kg}}\)
= 0.5 mol kg^-1
ΔT_f = i × K_f × m or i = \(\frac{\Delta \mathrm{T}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{b}} \times \mathrm{m}}\)
(or) i = \(\frac{1(\mathrm{~K})}{1.86 \mathrm{k} \mathrm{kg} \mathrm{mol}^{-1} \times 0.5 \mathrm{~mol} \mathrm{~kg}^{-1}}\)
= 1.0753
Each molecule of CH₂F COOH dissociate into 2 particles as
CH₂F COOH \(\rightleftharpoons\) CH₂F COO^– + H⁺

Question 48.
100g of liquid A(molar mass 140g mol^-1) was dissolved in 1000g of liquid B(molar mass 180g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.
Answer:
Let vapour pressure of pure A be \(\mathbf{p}_{\mathrm{A}}^0\)
\(p_{\mathrm{B}}^0\) = 500 torr ; n_A = \(\frac{100}{140}\) = 0.714 mol
n_B = \(\frac{1000}{180}\) = 5.55 mol
x_A = \(\frac{0.714}{5.55+0.714}\) = 0.114
x_B = 1 – 0.114 = 0.886
p_total = \(p_A^0 x+p_B^0 x_B\)
(or) 475(torr) = \(p_A^0\) × 0.114 + 500 × 0.886
(or) \(\mathrm{p}_{\mathrm{A}}^0\) = 280.7 torr
P_A = 280.70 × 0.114 = 32 Torr

Question 49.
Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
Answer:
We know that π = CRT
or π = \(\frac{\mathrm{i} \times \mathrm{W}_{\mathrm{B}} \times \mathrm{R} \times \mathrm{T}}{\mathrm{M}_{\mathrm{B}} \times \mathrm{V}_{\mathrm{L}}}\)
or W_B = \(\frac{\pi \times M_B \times V_L}{i \times R \times T}\)
van’t Hoff factor i = 2.47
Osmotic pressure p = 0.75 atm.
temperature T = 273 + 27 = 300 K.
Mol. wt of CaCl₂, M_B = 111
Volume of solution = 2.5 L
Constant R = 0.0821 L atm mol^-1 K^-1
W_B = \(\frac{0.75 \times 111 \times 2.5}{2.7 \times 0.0821 \times 300}\) = 3.40 g
Moles of CaCl₂ = \(\frac{3.4}{111}\) = 0.03 mol.

Question 50.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in two litre of water at 25°C assuming that it is completely disassociated.
Answer:
K₂SO₄ is strong electrolyte, ionises completely
K₂SO₄ → 2K⁺ + \(\mathrm{SO}_4^{2-}\)
Therefore i = 3
Volume of the solution = 2L
Mass of K₂SO₄ (W_B) = 25 mg = 0025 g
Molar mass of K₂SO₄ (M_B) = 174 g mol^-1
π = iCRT = i × \(\frac{3 \times 0.025 \times 0.0821 \times 298}{174 \times 2}\)
= 5.27 × 10^-3 atm

Question 51.
Benzene and Toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300K are 50.71 mm of Hg and 32.06 mm of Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80g of benzene is mixed with 100g of toluene.
Answer:
If y_b is the mole fraction of benzene in vapour form

Intext Questions – Answers

Question 1.
Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22g of benzene is dissolved in 122g of carbon tetrachloride.
Answer:
Weight of Benzene, w = 22 g
Weight of carbon tetrachloride, W = 122 g
Mass percentage of benzene = \(\frac{w \times 100}{W+w}\)
= \(\frac{22 \times 100}{22+122}\) = 15.27%
Mass percentage of carbon tetrachloride
= \(\frac{W \times 100}{W+w}\) = \(\frac{122 \times 100}{22+122}\) = 84.73%

Question 2.
Calculate the mole fraction of benzene in solution containing 30% by mass of it in carbon tetrachloride.
Answer:
Mass of benzene, w = 30 g
Mol. wt of benzene = 78 g mol^-1
Moles of benzene = \(\frac{30}{78}\) = 0.385
Mass of CCl₄ = 70 g
Mol. wt. of CCl₄ = 154 g mol^-1
Moles of CCl₄ = \(\frac{70(\mathrm{~g})}{154(\mathrm{~g})}\) = 0.454
Mole fraction of benzene = \(\frac{0.385}{0.385+0.454}\)
Mole fraction of CCl₄ = (1 – X_Benzene)
= 1 – 0.459 = 0.541

Question 3.
Calculate the molarity of each of the following solutions
(a) 30 g of CO(NO₃)₂. 6H₂O in 4.3L of solution
(b) 30 ml of 0.5 M H₂SO₄ diluted to 500 ml.
Answer:
a) Mass of CO (NO₃)₂. 6H₂0 = 30 g
Molar mass of CO(NO₃)₂. 6H₂O
= 297 g mol^-1.
Volume of solution = 4.3 L
Molarity M = \(\frac{30(\mathrm{~g})}{297 \mathrm{gmol}^{-1} \times 4.3 \mathrm{~L}}\)
= 0.023 mol L^-1

b) Initial volume V₁ = 30 ml
Molarity M₁ = 0.5 mol L^-1
Final volume V₂ = 500 ml
Final molarity M₂ =?
M₁V₁ = M₂V₂ (or) M₂ = \(\frac{M_1 V_1}{V_2}\)
= \(\frac{0.5 \times 30}{500}\) = 0.03 mol L^-1

Question 4.
Calculate the mass of urea (NH₂CONH₂) required in makIng 2.5 kg of 0.25 molal aqueous solution.
Answer:
The mass of urea required = W_B
Molar mass of urea M_B = 60
Mass of solution W_A + W_B = 2.5 kg = 2500 gm.
Mass of solvent W_A = 2500 – W_B
Molarity m = 0.25 mol kg^-1
m = \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{A}} \times \mathrm{W}_{\mathrm{B}}}\)
0.25 = \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{6.0 \times\left(2500-\mathrm{W}_{\mathrm{B}}\right)}\)
Solving we get W_B = 37.5 g.

Question 5.
Calculate
(a) molality
(b) molarity
(c) mole fraction of KI if the density of 20% (mass/ mass) aqueous KI is 1.202 g mL^-1
Answer:
Mass of KI(W_B) = 20 g
Molar mass of KI (M_B) = 166 g mol^-1
Density of solution (d) = 1.202 g mL^-1
Mass of solution (W_A + W_B) = 100 g
Volume of solution (V) = \(\frac{\text { Mass }}{\text { density }}\)
= \(\frac{100}{1.202 \mathrm{gmL}^{-1}}\) = 83.2 ml
Mass of solvent W_A = 100 – 20 = 80 g.

a) Molality (m) = \(\frac{W_B \times 1000}{M_B \times W_A}\)
= \(\frac{20 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{166 \mathrm{~g} \mathrm{~mol}^{-1} \times 80 \mathrm{~g}}\)
= 1.51 mol kg^-1

b) Molarity (M) = \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{V}}\)
= \(\frac{20 \mathrm{~g} \times 1000 \mathrm{~mL}}{166 \mathrm{~g} \mathrm{~mol}^{-1} \times 83.2 \mathrm{ml}}\)
= 1.45 mol L^-1

c)
Moles of KI(n_B) = \(\frac{20 \mathrm{~g}}{166 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 1.51 mol
Moles of water (n_A) = \(\frac{1000 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 55.5
∴ Mole fraction of KI (x_B) = \(\frac{1.51}{1.51+55.5}\)
= 0.026

Question 6.
H₂S, a toxic gas with rotten egg like smell, is used for qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.
Answer:
According to Henry’s law
K_H . x = p or K_H = \(\frac{p}{x}\)
P = 1 bar, solubility = 0.195 mol kg^-1
Moles of H₂S (n_B) = 0.195
Moles of water (n_A) = \(\frac{1000 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 55.5
∴ Mole fraction of H₂S (x)
= \(\frac{0.195}{0.195+55.5}\) = 3.5 × 10^-3
K_H = \(\frac{1(\text { bar) }}{3.5 \times 10^{-3}}\) = 285.6 bar

Question 7.
Henry’s law constant of CO₂ in water is 1.67 × 10⁸ Pa at 298K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 25 atm CO₂ pressure at 298 K.
Answer:
Pressure of CO₂ (p) = 2.5 atm
= 2.5 atm × 101325\(\frac{(\mathrm{Pa})}{\mathrm{atm}}\)
= 253312.5 Pa
According to Henrys law

Moles of CO₂ present in 500 ml
= \(\frac{0.0844}{2}\) mol = 0.0422 mol.
Amount of CO₂ present in 500 ml
= 0.0422 mol × 440 g mol^-1 = 1.86 g.

Question 8.
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively at 350K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Answer:
Total vapour pressure P_T = 600 mm Hg
Let Mole fraction of A is X_A and that of B is X_B
Vapour pressure of pure A(\(\mathrm{P}_{\mathrm{A}}^0\)) = 450 mm Hg
Vapour pressure of pure B(\(\mathrm{P}_{\mathrm{B}}^0\)) =700 mm Hg
Now P_Total = \(\mathrm{P}_{\mathrm{A}}^0 \mathrm{x}_{\mathrm{A}}\) + \(\mathrm{P}_{\mathrm{B}}^0 \mathrm{x}_{\mathrm{B}}\)
∴ 600 = 450 x_A + 700 x_B
= 450x_A + 700(1 – x_A)
= 700 – 250 x_A or x_A = 0.4
x_B = 1 – x_A = 1 – 0.4 = 0.6
Mole fraction of A in vapour phase
Y_A = \(\frac{x_A P_A^0}{P_{\text {Total }}^0}\) = \(\frac{0.4 \times 450}{600}\) = 0.3
Mole fraction of B in vapour phase
Y_B = \(\frac{\mathrm{x}_{\mathrm{B}} \mathrm{P}_{\mathrm{B}}^0}{\mathrm{P}_{\text {Total }}}\) = \(\frac{0.6 \times 700}{600}\) = 0.7

Question 9.
Vapour pressure of pure water at 298K is 23.8 mm Hg. 50 of urea (NH₂ CO NH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Answer:
Mass of water = 850 g
Moles of water = \(\frac{850}{18}\) = 47.22 mol
Mass of urea = 50 g
Mol mass of urea = 60
Moles of urea = \(\frac{50}{60}\) = 0.83
Since the solution is not dilute the formula to be used is

Question 10.
Boiling point of water at 750 min Hg is 99.63°C. How much sucrose is to be added to 500g of water such that it boils at 100°C.
Answer:
Let mass of sucrose required = W_B g
Mass of solvent W_A = 500 g
Molar mass of sucrose (C₁₂H₂₂O₁₁) = 342
Elevation in boiling point ΔT_b = 100 – 99.63
= 0.37°C
ΔT_b = \(\frac{\mathrm{K}_{\mathrm{b}} \times \mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{~W}_{\mathrm{A}} \times \mathrm{M}_{\mathrm{B}}}\)
(or) W_B = \(\frac{\Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{W}_{\mathrm{A}} \times \mathrm{M}_{\mathrm{B}}}{\mathrm{K}_{\mathrm{b}} \times 1000}\)
= \(\frac{0.37 \times 500 \times 342}{0.52 \times 1000}\) = 121.

Question 11.
Calculate the mass of ascorbic acid (vitamin C, C₆H₈O₆) to be dissolved in 75g of
acetic acid to lower its melting point by 1.5°C. K_f = 3.9 K kg mol^-1.
Answer:
Let the mass of ascorbic acid required = W_Bg
Molar mass of ascorbic acid (C₆H₈O₆) M_B = 176 g mol^-1
Mass of solvent (W_A) = 75 g
Depression in melting point ΔT_f = 1.5°C

Question 12.
Calculate the osmotic pressure In pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 ion 450 ml of water at 37°C.
Answer:
Mass of polymer (W_B) = 1.0 g
Molar mass of polymer M_B = 1,85,000 g mol^-1
T = 37 + 273.15 = 310.15K
π = \(\frac{W_B R_T}{M_B V_{(L)}}\)
= \(\frac{1 \times 0.082 \times 310.15}{185000 \times 450 \times 10^{-3}}\)
= 3.05 × 10^-4 atm
= 3.05 × 10^-4 × 1.01 × 10⁵Pa = 30.9 Pa

Problems

Question 1.
Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂ by mass. (TS 15)
Answer:
Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20 g of ethylene glycol and 80 g of water.
Molar mass of C₂H₆O₂ = 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol^-1.

Question 2.
Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution. (TS Mar. 19; AP & TS 15)
Answer:
Moles of NaOH = \(\frac{5 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.125 mol
Volume of the solution in litres = 450 mL /1000 mL L^-1
Using equation (2.8)
Molarity = \(\frac{0.125 \mathrm{~mol} \times 1000 \mathrm{~mL} \mathrm{~L}^{-1}}{450 \mathrm{~mL}}\)
= 0.278 M = 0.278 mol L^-1
= 0.278 mol dm^-3

Question 3.
Calculate molality of 2.5 g of ethanoic acid (CH₃COOH) in 75 g of benzene. (TS 15)
Answer:
Molar mass of C₂H₄O₂ = 12 × 2 + 1 × 4 + 16 × 2 = 60 g mol^-1.
Moles mass of C₂H₄O₂ = \(\frac{2.5 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.0417 mol
Mass of benzene in kg = 75 g / 1000 g kg^-1
= 75 × 10^-3 kg
Molality of C₂H₄O₂ = \(\frac{\text { moles of } \mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2}{\text { kg of benzene }}\)
= \(\frac{0.0417 \mathrm{~mol} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{75 \mathrm{~g}}\)
= 0.556 mol kg^-1

Question 4.
If N₂ gas is bubbled through water at 293K, how many mililmoles of N₂ gas would dissolve in 1 litre of water? Assume that N₂ exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N₂ at 293 K is 76.48 kbar.
Answer:
The solubility of gas is related to the mole fraction in aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry’s law. Thus:

As 1 litre of water coñtains 55.5 mol of it, therefore If n represents number of moles of N₂ in solution.
x(nitrogen) = \(\frac{n \mathrm{~mol}}{\mathrm{n} \mathrm{mol}+55.5 \mathrm{~mol}}\)
= \(\frac{\mathrm{n}}{55.5}\) = 1.29 × 10^-5
(n in denominator is neglected as it is <<55.5)
Thus n = 1.29 × 10^-5 × 55.5 mol
= 7.16 × 10^-4 mol^-1
= \(\frac{7.16 \times 10^{-4} \mathrm{~mol} \times 1000 \mathrm{~m} \mathrm{~mol}}{1 \mathrm{~mol}}\)
= 0.716 m mol

Question 5.
Vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 298 K are 200mm Hg and 415 mm Hg respectively.
(i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 298 K and,
(ii) mole fractions of each component in vapour phase.
Answer:
i) Molar mass of CH₂Cl₂ = 12 × 1 + 1 × 2 + 35.5 × 2 = 85 g mol^-1
Molar mass of CHCl₃ = 12 × 1 + 1 × 1 + 35.5 × 3 = 119.5 g mol^-1
Moles of CH₂Cl₂ = \(\frac{40 \mathrm{~g}}{85 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.47 mol
Moles of CHCl₃ = \(\frac{25.5 \mathrm{~g}}{119.5 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.213 mol
Total number of moles = 0.47 + 0.213
= 0.683 mol

Using equation (2.16),
P_total = \(\mathrm{p}_1^0\) + (\(\mathrm{p}_2^0\) + \(\mathrm{p}_1^0\))x₂
= 200 + (415 – 200) × 0.688
= 200 + 147.9 = 347.9 mm Hg

(ii) Using the relation (2.19) y₁ = P₁ / p_total, we can calculate the mole fraction of the components in gas phase (y₁).
\(\mathrm{p}_{\mathrm{CH}_2} \mathrm{Cl}_2\) = 0.688 × 415 mm Hg = 285.5 mm Hg
\(\mathbf{p}_{\mathrm{CHCl}_3}\) = 0.312 × 200 mm Hg = 62.4 mm Hg
\(\mathbf{y}_{\mathrm{CH}_2 \mathrm{Cl}_2}\) = 285.5 mm Hg / 347.9 mm Hg = 0.82
\(\mathbf{y}_{\mathrm{CHCl} 3}\) = 62.4 mm Hg/ 347.9 mm Hg = 0.18