{"id":37762,"date":"2022-12-03T16:59:35","date_gmt":"2022-12-03T11:29:35","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=37762"},"modified":"2022-12-03T17:04:02","modified_gmt":"2022-12-03T11:34:02","slug":"ts-inter-2nd-year-chemistry-study-material-chapter-2","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-chemistry-study-material-chapter-2\/","title":{"rendered":"TS Inter 2nd Year Chemistry Study Material Chapter 2 Solutions"},"content":{"rendered":"

Telangana TSBIE\u00a0TS Inter 2nd Year Chemistry Study Material<\/a> 2nd Lesson Solutions Textbook Questions and Answers.<\/p>\n

TS Inter 2nd Year Chemistry Study Material 2nd Lesson Solutions<\/h2>\n

Very Short Answer Questions (2 Marks)<\/span><\/p>\n

Question 1.
\nDefine the term solution.
\nAnswer:
\nHomogeneous mixture of two or more than two components is called solution.<\/p>\n

Question 2.
\nDefine molarity.
\nAnswer:
\nMolarity (M) is defined as number of moles of solute dissolved in one litre of (or one cubic decimetre) solution.
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\nDefine molality.
\nAnswer:
\nMolality (m) is defined as the number of moles of the solute present in one kilogram (kg) of the solvent.
\n\"TS<\/p>\n

Question 4.
\nGive an example of a solid solution in which the solute is solid.
\nAnswer:
\nIn solid solution solute is solid. Here sol-vent is also solid. e.g., copper dissolved in gold (alloys).<\/p>\n

Question 5.
\nDefine mole fraction. (Mar. 2018 TS)
\nAnswer:
\nMole fraction is the ratio of number of moles of one component to the total number of moles of all the components in the solution.
\nMole fraction of a component
\n\"TS<\/p>\n

Question 6.
\nDefine mass percentage solution.
\nAnswer:
\nThe mass percentage of a component of a solution is defined as
\nMass % of a component
\nMass % of the component
\n\"TS<\/p>\n

Question 7.
\nWhat is ppm of a solution ?
\nAnswer:
\nParts per million
\nNumber of parts of the
\n\"TS
\nThis method is convenient when a solute is present in trace amounts.<\/p>\n

Question 8.
\nWhat role do the molecular interactions play in a solution of alcohol and water ?
\nAnswer:
\nThe interaction between alcohol and alcohol is hydrogen bonding and H2<\/sub>O and H2<\/sub>O is also hydrogen bonding. After mixing the interaction between alcohol and H2<\/sub>O is also hydrogen bonding but less than the hydrogen bonding in alcohol and H2<\/sub>O separately. So after mixing the interaction between alcohol and water decreases and the vapour pressure of solution increases forming low boiling point azeotrope.<\/p>\n

Question 9.
\nState Raoult\u2019s law. (AP & TS 16) (IPE 14) (Mar. 2018 . AP & TS)
\nAnswer:
\nFor any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.<\/p>\n

Question 10.
\nState Henry’s law.
\nAnswer:
\nThe 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.
\n(or)
\nMole fraction of gas in the solution is proportional to the partial pressure of the gas over the solution.
\n(or)
\nThe partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.
\nP = KH<\/sub> \u00d7 x
\nKH<\/sub> is the Henry’s law constant.<\/p>\n

Question 11.
\nWhat is Ebullioscopic constant?
\nAnswer:
\nThe 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.
\n\u0394Tb<\/sub> = Kb<\/sub> m.
\nunit of Kb<\/sub> = K kg mol-1<\/sup><\/p>\n

Question 12.
\nWhat is Cryoscopic constant?
\nAnswer:
\nThe 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.
\n\u0394Tf<\/sub> = Kf<\/sub> m.
\nunit of Kf<\/sub> = K kg mol-1<\/sup><\/p>\n

\"TS<\/p>\n

Question 13.
\nDefine Osmotic pressure. (AP 16, 15) (Mar. 2018 – AP)
\nAnswer:
\nWhen 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.
\n(or)
\nThe 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.<\/p>\n

Question 14.
\nWhat are isotonic solutions ? TS Mar. 19; (AP 17, 15; TS 16)
\nAnswer:
\nTwo solutions having same osmotic pressure at a given temperature are called isotonic solutions.<\/p>\n

Question 15.
\nAmongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water,
\n(i) Phenol
\n(ii) Toluene
\n(iii) Formic acid
\n(iv) Ethylene glycol
\n(v) Chloroform
\n(vi) Pentanol.
\nAnswer:
\nInsoluble : Chloroform, toluene.
\nPartially soluble : Phenol, pentanol.
\nHighly soluble: Formic acid, ethylene glycol.<\/p>\n

Question 16.
\nCalculate the mass percentage of aspirin (C9<\/sub>H8<\/sub>O4<\/sub>) in acetonitrile (CH3<\/sub>CN) when 6.5 gm of C9<\/sub>H8<\/sub>O4<\/sub> is dissolved in 450g of CH3<\/sub>CN.
\nAnswer:
\nMass percentage of aspirin.
\n\"TS<\/p>\n

Question 17.
\nCalculate the amount of benzoic acid (C6<\/sub>H5<\/sub>COOH) required for preparing 250ml of 0.15M solution in methanol.
\nAnswer:
\nMolecular mass of benzoic acid (C6<\/sub>H5<\/sub>COOH) = 12 \u00d7 6 + 5 + 12 + 32 + 1 = 122 g mol-1<\/sup>
\nmoles of benzoic = M \u00d7 V
\n= 250 \u00d7 0.15 = 37.5
\n= 37.5 \u00d7 10-3<\/sup> mol
\nAmount of benzoic acid = Moles \u00d7 MW2<\/sub>
\n= 37.5 \u00d7 10-3<\/sup> \u00d7 122
\n= 4.575 g.<\/p>\n

Question 18.
\nThe 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.
\nAnswer:
\nIncreasing order of \u0394Tf<\/sub> (depression in freezing point) is Acetic acid < dichloroacetic acid < trichloroacetic acid.
\nDue 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.<\/p>\n

Question 19.
\nWhat is van’t Hoffs factor ‘i’ and how is it related to ‘\u03b1’ in the case of a binary ele-ctrolyte (1 : 1) ?
\nAnswer:
\nvan’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
\n\"TS
\nIf \u03b1 represents the degree of association of the solute for an equilibrium.
\nnA \\(\\rightleftharpoons\\) An
\nTotal number of particles at equilibrium is 1 – \u03b1 + \\(\\frac{\\alpha}{2}\\) = 1 – \\(\\frac{\\alpha}{2}\\)
\nThen i = \\(\\frac{\\alpha \/ 2}{1 \\cdot 1}\\) = \\(\\frac{\\alpha}{2}\\)
\nFor dissociation
\n\"TS
\nTotal moles of particles aren(1 – \u03b1 + \u03b1 + \u03b1)
\n= n(1 + \u03b1)
\ni = \\(\\frac{n(1+\\alpha)}{n}\\) = 1 + \u03b1<\/p>\n

\"TS<\/p>\n

Question 20.
\nWhat is relative lowering of vapour pressure ? (AP Mar. 19)
\nAnswer:
\nThe 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.
\nIt can be shown as R.L.V.P. = \\(\\frac{p^{\\circ}-p}{p^{\\circ}}\\)
\nHere p\u00b0 is the vapour pressure of pure solvent, p is the vapour pressure of solution containing non – volatile solute.<\/p>\n

Question 21.
\nCalculate the mole fraction of H2<\/sub>SO4<\/sub> in a solution containing 98% H2<\/sub>SO4<\/sub> by mass. (Mar. 2018 – TS)(IPE 14)
\nAnswer:
\nMass of water = 2 gm.
\nMoles of water = \\(\\frac{2}{18}\\) = \\(\\frac{1}{9}\\) (M.Wt. of H2<\/sub>O = 18)
\nMass of H2<\/sub>SO4<\/sub> = 98 gm.
\nMoles of H2<\/sub>SO4<\/sub> = \\(\\frac{98}{98}\\)
\n= 1 (Mol. Wt. of H2<\/sub>SO4<\/sub> = 98)
\nMoles fraction of H2<\/sub>SO4<\/sub> = \\(\\frac{1}{1+\\frac{1}{9}}\\) = \\(\\frac{9}{10}\\) = 0.9<\/p>\n

Question 22.
\nHow many types of solutions are formed ? Give an example for each type of solution.
\nAnswer:
\nDepending on the type of solvent three types of solutions will be formed.<\/p>\n\n\n\n\n\n\n
Type of Solution<\/td>\nSolute<\/td>\nSolvent<\/td>\nCommon Examples<\/td>\n<\/tr>\n
Gaseous Solutions<\/td>\nGas
\nLiquid
\nSolid<\/td>\n		Gas
\nGas
\nGas<\/td>\n		Mixture of oxygen and nitrogen gases
\nChloroform mixed with nitrogen gas
\nCamphor in nitrogen gas.<\/td>\n<\/tr>\n
Liquid Solutions<\/td>\nGas
\nLiquid
\nSolid<\/td>\n		Liquid
\nLiquid
\nLiquid<\/td>\n		Oxygen dissolved in water.
\nEthanol dissolved in water.
\nGlucose dissolved in water.<\/td>\n<\/tr>\n
Solid Solutions<\/td>\nGas
\nLiquid
\nSolid<\/td>\n		Solid
\nSolid
\nSolid<\/td>\n		Solutions of hydrogen in palladium.
\nAmalgam of mercury with sodium.
\nCopper dissolved in gold.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

 <\/p>\n

Question 23.
\nDefine mass percentage, volume percentage and mass to volume percentage solutions.
\nAnswer:
\nMass percentage (w\/w) : The mass percentage of a component of a solution is defined as
\nMass % of a component
\nMass of the component
\n\"TS
\nVolume percentage (V\/V) : The volume percentage is defined as
\nVolume % of a component
\n\"TS
\nMass to volume percentage (w\/v): It is the mass of solute dissolved in 100 ml of the solution.<\/p>\n

Question 24.
\nConcentrated 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<\/sup>?
\nAnswer:
\n68% by mass implies that 68 g of HNO3<\/sub> is present in 100 g of solution.
\n\"TS<\/p>\n

Question 25.
\nA solution of glucose in water is labelled as 10% w\/w. What would be the molarity of the solution?
\nAnswer:
\nWt. of glucose = 1o g
\nWt. of solution = 100 gm
\nWt. of water = 100 – 10 = 90 gm.
\nConsidering density of water as 1 gm mL-1<\/sup>
\nVolume of water = 90 mL
\n\"TS<\/p>\n

Question 26.
\nA solution of sucrose In water is labelled as 20% w\/w. What would be the mole fraction of each component in the solution?
\nAnswer:
\nWt. of sucrose = 20
\nMoles of sucrose = \\(\\frac{20}{342}\\) = 0.0585
\nWt. of solution = 100 g
\nWt. of water = 100 – 20 = 80 g
\nMoles of water = \\(\\frac{80}{18}\\) = 4.44
\nMole fraction of sucrose = \\(\\frac{0.0585}{0.0585+4.44}\\)
\n= 0.013
\nMole fraction of water = \\(\\frac{4.44}{0.0585+4.44}\\)
\n= 0.987<\/p>\n

\"TS<\/p>\n

Question 27.
\nHow many ml of 0.1 MHCl is required to react completely with 1.0 g mixture of Na2<\/sub>CO3<\/sub> and NaHCO3<\/sub> containing equimolar amounts of both ?
\nAnswer:
\nLet Na2<\/sub>CO3<\/sub> be x g
\n\"TS
\nSince, given mixture contains equimolar amount.
\n\u2234 \\(\\frac{\\mathrm{x}}{106}\\) = \\(\\frac{1-x}{84}\\) (or) x = 0.56
\nTotal HCl required
\n= \\(\\left[\\frac{2 x}{106}+\\frac{1-x}{84}\\right]\\) = \\(\\frac{2 \\times 0.56}{106}\\) + \\(\\frac{1-0.56}{84}\\)
\n= 0.01578 mol
\nIf V is the volume of HCl required, then
\nV(L)<\/sub> \u00d7 0.1 = 0.01578
\n(or) V(L)<\/sub> = 0.1578L (or) 157.8 mL.<\/p>\n

Question 28.
\nA solution is obtained by mixing 300g of 25% solution and 400g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
\nAnswer:
\nMass of solute in 300 g of 25% solution.
\n= \\(\\frac{300 \\times 25}{100}\\) = 75 g
\nMass of solute in 400 g of 40% solution
\n= \\(\\frac{400 \\times 40}{100}\\) = 160 g
\nTotal mass of solute = 75 + 160 = 235 g
\nTotal mass of solution = 700 g
\n% solute in final solution = \\(\\frac{235 \\times 100}{700}\\)
\n= 33.5%
\n% of water in final solution = 100 – 33.5
\n= 66.5%<\/p>\n

Question 29.
\nAn antifreeze solution is prepared from 222.6g of ethylene glycol (C6<\/sub>H6<\/sub>O2<\/sub>) and 200g of water (solvent). Calculate the molality of the solution.
\nAnswer:
\nMass of glycol (w1<\/sub>) = 222.6 g
\nMoles of glycol (n1<\/sub>)<\/p>\n

= \"TS
\n= \\(\\frac{222.6}{62}\\)
\n= 3.59 mol
\nMass of water (w2<\/sub>) = 200 g = \\(\\frac{200 \\times 1 \\mathrm{~kg}}{1000}\\)
\n= 0.2 kg
\nMass of solution (w1<\/sub> + w2<\/sub>) = 200 + 222.6
\n= 422.6 g
\nDensity of solution (d) = 1.072 g mL-1<\/sup>
\nVolume of solution (V)
\n= \"TS
\n= \\(\\frac{422.6}{1.072}\\)
\n= 394.22 mL (or) 0.3942 L
\nMolality (m) = \\(\\frac{\\mathrm{n}_1}{\\mathrm{w}_2 \\mathrm{~kg}}\\) = \\(\\frac{3.59 \\mathrm{~mol}}{0.2 \\mathrm{~kg}}\\)
\n= 17.95 mol kg
\nMolarity (M) = \\(\\frac{n_1}{V \\text { in lit }}\\) = \\(\\frac{3.59}{0.3942}\\) = 9.1 mol L-1<\/sup><\/p>\n

Question 30.
\nWhy do gases always tend to be less soluble in liquids as the temperature is raised ?
\nAnswer:
\nDissolution of gases is exothermic process. It is because dissolution of a gas in a liquid decreases the entropy (\u0394S < 0). Thus increase of temperature tends to push the equilibrium (Gas + solvent \\(\\rightleftharpoons\\) solution, \u0394H = -Ve) in the backward direction, thereby, suppressing the dissolution.<\/p>\n

\"TS<\/p>\n

Question 31.
\nWhat is meant by positive deviations from Raoult’s law and how is the sign of \u0394mix<\/sub> H related to positive deviation from Raoult’s law ?
\nAnswer:
\nThe 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.<\/p>\n

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.
\n\"TS
\nFor this type of non – ideal solutions exhibiting positive deviations.<\/p>\n

    \n
  1. PA<\/sub> < \\(P_A^0 x_A\\); PB<\/sub> > \\(\\mathrm{P}_{\\mathrm{B}}^0 \\mathrm{x}_{\\mathrm{B}}\\)<\/li>\n
  2. \u0394mix<\/sub>H = +ve<\/li>\n
  3. \u0394mix<\/sub>V = +ve<\/li>\n<\/ol>\n

    Question 32.
    \nWhat is meant by negative deviation from Raoult’s law and how is the sign of \u0394mix<\/sub>H related to negative deviation from Raoult\u2019s law?
    \nAnswer:
    \nThe solutions which do not obey Raoult\u2019s 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\u2019s law. In this type of solutions solvent – solvent and solute – solute interacti\u00f2ns are weaker than that of solvent – solute interactions.<\/p>\n

    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\u2019s law.
    \n\"TS<\/p>\n

    For these solutions exhibiting negative deviations<\/p>\n

      \n
    1. \"TS<\/li>\n
    2. \u0394mix<\/sub> H = -ve<\/li>\n
    3. \u0394mix<\/sub>V = -ve<\/li>\n<\/ol>\n

      Question 33.
      \nThe vapour pressure of water is 12.3 k Pa<\/sub> at 300K. Calculate the vapour pressure of 1 molar solution of a non – volatile solute in it.
      \nAnswer:
      \nVapour pressure of water \\(\\mathrm{P}_{\\mathrm{H}_2}^0 \\mathrm{O}\\) 12.3 k Pa
      \nIn 1 molar solution
      \nMoles of water nH2<\/sub>O<\/sub> = \\(\\frac{1000}{18}\\) = 55.5 mol
      \nMoles of solute nB<\/sub> = 1 mol
      \nMole fraction of H2<\/sub>O (XH2<\/sub>O<\/sub>) = \\(\\frac{\\mathrm{n}_{\\mathrm{H}_2 \\mathrm{O}}}{\\mathrm{n}_{\\mathrm{H}_2 \\mathrm{O}}+\\mathrm{n}_{\\mathrm{B}}}\\)
      \n= \\(\\frac{55.5}{55.5+1}\\) = 0.982
      \n\"TS<\/p>\n

      Question 34.
      \nCalculate the mass of a non – volatile solute (molar mass 40g mol-1<\/sup>) which should be dissolved in 114g Octane to reduce its vapour pressure to 80%. (TS 16; IF\u2019 14)
      \nAnswer:
      \nVapour pressure of solution (p) = 80% of \\(\\mathrm{P}_{\\mathrm{A}}^0\\) = 0.8\\(\\mathrm{P}_{\\mathrm{A}}^0\\)
      \nLet the mass of solute be w g
      \n\u2234 Moles of solute nB<\/sub> = \\(\\frac{\\mathrm{W}_{\\mathrm{B}}}{\\mathrm{M}_{\\mathrm{B}}}\\) = \\(\\frac{W}{40}\\) mol
      \n\"TS<\/p>\n

      Question 35.
      \nA 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.
      \nAnswer:
      \nMolarity of sugar solution (m) = \\(\\frac{\\mathrm{W}_{\\mathrm{B}} \\times 1000}{\\mathrm{M}_{\\mathrm{B}} \\times \\mathrm{W}_{\\mathrm{A}}}\\)
      \n(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<\/sup>
      \n\u0394Tf<\/sub> = 273.15 – 271 = 2.15\u00b0
      \nNow \u0394Tf<\/sub> = Kf<\/sub> \u00d7 m or Kf<\/sub> = \\(\\frac{\\Delta \\mathrm{T}_{\\mathrm{f}}}{\\mathrm{m}}\\) = \\(\\frac{2.15}{0.154}\\) = 13.96
      \nMolailty of glucose solution (m)
      \n= \\(\\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<\/sup>
      \n\u0394Tf<\/sub> = Kf<\/sub>m= 13.96 \u00d7 0.292 = 4.08\u00b0
      \nFreezing point of glucose solution
      \n= 273.15 – 4.08 = 269.07 K.<\/p>\n

      Question 36.
      \nIf the osmotic pressure of glucose solution is 1.52 bar at 300K. What would be its concentration if R = 0.083L bar mol-1<\/sup> K-1<\/sup>?
      \nAnswer:
      \nTemperature T = 300 K
      \nOsmotic pressure = 1.52 bar
      \nR = 0.083 L bar mol-1<\/sup> K-1<\/sup>
      \n\u03c0 = CRT (or) C = \\(\\frac{\\pi}{\\mathrm{RT}}\\)
      \n= \\(\\frac{1.52 \\mathrm{bar}}{0.083 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1} \\times 300}\\)
      \n\u2234 = 0.061 mol L-1<\/sup><\/p>\n

      Question 37.
      \nVapour 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)
      \nAnswer:
      \nVapour pressure of water \\(\\mathrm{p}_{\\mathrm{H}_2 \\mathrm{O}}^{\\mathrm{O}}\\) = 17.535 mm.
      \nLet vapour pressure of solution be ps<\/sub>
      \n\"TS<\/p>\n

      Question 38.
      \nHow is molar mass related to the elevation in boiling point of a solution ?
      \nAnswer:
      \nTo calculate the molar mass of an unknown non – volatile compound a known mass (say WB<\/sub>g) of it is dissolved in a known mass (say WA<\/sub> g) of some suitable solvent and elevation in its boiling point (\u0394Tb<\/sub>) is determined. Let MB<\/sub> be the molar mass of the compound. Then
      \nMolarity of solution m = \\(\\frac{W_B}{M_B} \\times \\frac{1000}{W_A}\\)
      \nWe know, \u0394Tb<\/sub> = Kb<\/sub> \u00d7 m = Kb<\/sub> \u00d7 \\(\\frac{W_B}{W_A}\\) \u00d7 \\(\\frac{1000}{M_B}\\)
      \nMB<\/sub> = \\(\\frac{\\mathrm{K}_{\\mathrm{b}} \\times \\mathrm{W}_{\\mathrm{b}} \\times 1000}{\\mathrm{~W}_{\\mathrm{A}} \\times \\Delta \\mathrm{T}_{\\mathrm{b}}}\\)
      \nKnowing Kb<\/sub>, WB<\/sub>, WA<\/sub> and the molar mass of the compound can be calculated
      \nfrom the above relation. This method is known as ebullioscopic method.<\/p>\n

      Question 39.
      \nWhat is an Ideal solution?
      \nAnswer:
      \nAn ideal solution may be defined as the solution which obeys Raoults law over the entire range of concentration and temperature.<\/p>\n

      According to Raoult’s law, the vapour pressure of a volatile component (pA<\/sub>) of the solutions is equal to the product of its mole fraction (xA<\/sub>) in solution and vapour pressure in pure state \\(\\mathrm{P}_{\\mathrm{A}}^0\\).
      \nPA<\/sub> = \\(\\mathrm{p}_{\\mathrm{A}}^0 \\mathrm{x}_{\\mathrm{A}}\\)<\/p>\n

      The formation of ideal solution neither involve any change in enthalpy nor in volume. An ideal solution,<\/p>\n

        \n
      1. should obey Raoults law i.e., pA<\/sub> = \\(p_A^0 x_A\\) and pB<\/sub> = \\(p_B^0 x_B\\)<\/li>\n
      2. \u0394mix<\/sub>H = 0<\/li>\n
      3. \u0394mix<\/sub>V = 0<\/li>\n<\/ol>\n

        In ideal solutions the solvent – solvent and solute – solute interactions are almost the same type as solvent – solute interactions.<\/p>\n

        \"TS<\/p>\n

        Question 40.
        \nWhat is relative lowering of vapour pressure ? How is it useful to determine the molar mass of a solute ?
        \nAnswer:
        \nWhen 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.<\/p>\n

        The ratio of lowering of vapour pressure to that vapour pressure of pure solvent is known as relative lowering of vapour pressure.
        \n\\(\\frac{\\Delta \\mathrm{p}}{\\mathrm{p}_{\\mathrm{A}}^0}\\) = Relative lowering of vapour pressure
        \n\u0394p = \\(\\mathrm{p}_{\\mathrm{A}}^0-\\mathrm{p}_{\\mathrm{A}}\\) = lowering of vapour pressure
        \n\\(\\mathrm{p}_{\\mathrm{A}}^0\\) = vapour pressure of pure solvent<\/p>\n

        Determination of molar mass of a solute:
        \nAccording to Raoult\u2019s law the relative lowering of vapour pressure i equal to the mole fraction of the solute.
        \n\\(\\frac{\\Delta \\mathrm{p}}{\\mathrm{p}_{\\mathrm{A}}^0}\\) = xB<\/sub>
        \nxB<\/sub> = mole fraction of solute,<\/p>\n

        If a known mass (WB<\/sub>) of the solute is dissolved in a known mass (WA<\/sub>) of solvent to prepare a dilute solution and the relative lowering of vapour pressure Is determined experimentally the molar mass of solvent (MA<\/sub>) is known, the molar mass of solute MB<\/sub> can be determined as follows.
        \n\"TS
        \nIn this equation all the parameters except MB<\/sub> are known and hence MB<\/sub> can be calculated.<\/p>\n

        Question 41.
        \nHow is molar mass related to the depre-ssion in freezing point of a solution ?
        \nAnswer:
        \nTo determine the molar mass of an unknown non – volatile compound a known mass (say WB<\/sub>g) of it is dissolved in a known mass (say WA<\/sub>g) of some suitable solvent and depression in the freezing point (\u0394Tf<\/sub>) is determined. Let MB<\/sub> be the molar mass of the compound. Then
        \nMolarity of the solution m = \\(\\frac{W_B}{M_B} \\times \\frac{1000}{W_A}\\)
        \nWe know \u0394Tf<\/sub> = Kf<\/sub> \u00d7 m = Kf<\/sub> \u00d7 \\(\\frac{W_B}{W_A}\\) \u00d7 \\(\\frac{1000}{M_B}\\)
        \nMB<\/sub> = \\(\\frac{\\mathrm{K}_{\\mathrm{b}} \\times \\mathrm{W}_{\\mathrm{B}} \\times 1000}{\\mathrm{~W}_{\\mathrm{A}} \\times \\Delta \\mathrm{T}_{\\mathrm{f}}}\\)
        \nKnowing Kf<\/sub>, WB<\/sub>, WA<\/sub> and \u0394Tf<\/sub>, the molar mass of the compound can be calculated from the above relation. This method is called cryoscopic method.<\/p>\n

        Long Answer Questions (8 Marks)<\/span><\/p>\n

        Question 42.
        \nAn 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 ?
        \nAnswer:
        \nThe vapour pressure of pure water
        \n\\(\\mathbf{p}_{\\mathrm{A}}^0\\) = 1 atm = 1.013 bar
        \nVapour pressure of solution (p) = 1.004 bar
        \nWB<\/sub> = 2g = WA<\/sub> + WB<\/sub> = 100g; WA<\/sub> = 98g
        \nNow \\(\\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}\\)
        \nor MB<\/sub> = \\(\\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)}\\)
        \n= 41.35 g. mol-1<\/sup><\/p>\n

        Question 43.
        \nHeptane and Octane form an ideal solution. At 373K the vapour pressure of the two liquid components are 105.2 kPa<\/sub> and 46.8 kPa<\/sub> respectively. What will be the vapour pressure of a mixture of 26.0 g heptane and 35g of octane ?
        \nAnswer:
        \nMoles of C7<\/sub>H16<\/sub>(nH<\/sub>) = \\(\\frac{W_{(H)}}{M_{(H)}}\\) = \\(\\frac{25}{100}\\) = 0.25 mol
        \nMoles of C8<\/sub>H18<\/sub> (nO<\/sub>) = \\(\\frac{W_{(0)}}{M_{(0)}}\\) = \\(\\frac{35}{114}\\) = 0.307 mol
        \nMole fraction of C7<\/sub>H16<\/sub> = (XH<\/sub>) = \\(\\frac{0.25}{0.25+0.307}\\)
        \n= \\(\\frac{0.25}{0.557}\\) = 0.466
        \nMole fraction of C8<\/sub>H18<\/sub> (X0<\/sub>) = 1 – 0.449
        \n= 0.534
        \nVapour pressure of heptane (pH)
        \n\\(\\mathrm{p}_{\\mathrm{H}}^{\\mathrm{O}}\\) \u00d7 XH<\/sub>
        \n= 105.2 \\(\\mathrm{Kp}_{\\mathrm{a}}\\) \u00d7 0.449
        \n= 47.2348 \\(\\mathrm{Kp}_{\\mathrm{a}}\\) \u2245 49.02 \\(\\mathrm{kp}_{\\mathrm{a}}\\)
        \nVapour pressure of octane (PO<\/sub>)
        \n= \\(P_{\\mathrm{H}}^O\\) \u00d7 XO<\/sub>
        \n= 46.8 \\(\\mathrm{kp}_{\\mathrm{a}}\\) \u00d7 0.551 = 24.99 \\(\\mathrm{kp}_{\\mathrm{a}}\\)
        \nTotal vapour pressure PTotal<\/sub> = PH<\/sub> + PO<\/sub>
        \n= 49.02 + 24.99 = 74.09 \\(\\mathrm{kp}_{\\mathrm{a}}\\)<\/p>\n

        Question 44.
        \nA 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
        \n(i) The molar mass of the solute and
        \n(ii) Vapour pressure of water at 298k.
        \nAnswer:
        \nMoles of solute (nB<\/sub>) = \\(\\frac{30}{M_B}\\)
        \nMoles of H2<\/sub>O(nH2<\/sub>O) = \\(\\frac{90}{18}\\) = 5 mol
        \nMole fraction of H2<\/sub>O
        \n\"TS
        \nAfter adding 18g(= 1 mol) of water to solution new mole fraction of water(\\(\\mathrm{X}_{\\mathrm{H}_2 \\mathrm{O}}^{\\prime}\\)) is
        \n\"TS<\/p>\n

        Question 45.
        \nTwo elements A and B form compounds having formula AB2<\/sub> and AB4<\/sub>. When dissolved in 20g of Benzene (C6<\/sub>H6<\/sub>), 1g of AB2<\/sub> lowers the freezing point by 2.3K whereas 1.0g of AB4<\/sub> lowers it by 1.3K. The molar depression constant for benzene is 5.1 K kg mol-1<\/sup>. Calculate atomic masses of A and B.
        \nAnswer:
        \nWe know
        \n\"TS
        \nLet the atomic weight of A = x
        \nLet the atomic weight of B = y
        \nx + 2y = 110.86
        \nx + 4y = 196.15
        \nSolving for x and y, we get
        \ny = 42.64, x = 25.58
        \n\u2234 Atomic weight of A = 25.8 u
        \nAtomic weight of B = 42.64 u<\/p>\n

        Question 46.
        \nCalculate the depression In the freezing point of water when 10g of CH3<\/sub>CH2<\/sub>CHClCOOH is added to 250g of water. Ka<\/sub> = 1.4 \u00d7 10-3<\/sup>, Kf<\/sub> = 1.86K kg mol-1<\/sup>.
        \nAnswer:
        \nCalculation of molarity of the solution:
        \nMass of the solution = 250 + 10 = 260 g
        \n\"TS<\/p>\n

        Calculation of van’t Hoffs factor (i) : Let degree of dissociation of acid be \u03b1 since acid is monobasic acid, therefore \u03b1 and Ka<\/sub> of acid are related as
        \n\u03b1 = \\(\\sqrt{K_{\\mathrm{a}} \/ \\mathrm{c}}\\) = \\(\\sqrt{\\frac{1.4 \\times 10^{-3}}{0.284}}\\) = 0.07
        \nVan’t Hoff factor and degree of dissociation are related as
        \n\u03b1 = \\(\\frac{i-1}{m-1}\\) = \\(\\frac{i-1}{2-1}\\) or i = 1 + \u03b1 = 1 + 0.07
        \n(or) i = 0.7
        \nCalculation of depression in freezing point \u0394Tf<\/sub>
        \n\u0394Tf<\/sub> = i \u00d7 Kf<\/sub> \u00d7 m = \\(\\frac{1.07 \\times 1.86 \\times 10 \\times 1000}{122.5 \\times 250}\\)
        \n= 0.649 = 0.65\u00b0C<\/p>\n

        \"TS<\/p>\n

        Question 47.
        \n19.5 g of CH2<\/sub>FCOOH is dissolved in 500g of water. The depression in freezing point of water observed is 1.0\u00b0C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
        \nAnswer:
        \nMass of solution = 500 + 19.5 = 519.5 g.
        \n\"TS
        \nMoles of CH2<\/sub>F COOH (nB<\/sub>) = \\(\\frac{W_B}{M_B}\\) = \\(\\frac{19.5}{78}\\)
        \n= 0.25 mol
        \nMolarity of solution (M) = \\(\\frac{n_B}{V_{(L)}}\\) = \\(\\frac{0.25}{0.462}\\)
        \n= 0.541 moL-1<\/sup>
        \nMass of water (WA<\/sub>) = 500 g = 0.5 kg
        \nMolarity of solution (m) = \\(\\frac{\\mathrm{n}_{\\mathrm{B}}}{\\mathrm{W}_{\\mathrm{A}} \\mathrm{kg}}\\) = \\(\\frac{0.25 \\mathrm{~mol}}{0.25 \\mathrm{~kg}}\\)
        \n= 0.5 mol kg-1<\/sup>
        \n\u0394Tf<\/sub> = i \u00d7 Kf<\/sub> \u00d7 m or i = \\(\\frac{\\Delta \\mathrm{T}_{\\mathrm{b}}}{\\mathrm{K}_{\\mathrm{b}} \\times \\mathrm{m}}\\)
        \n(or) i = \\(\\frac{1(\\mathrm{~K})}{1.86 \\mathrm{k} \\mathrm{kg} \\mathrm{mol}^{-1} \\times 0.5 \\mathrm{~mol} \\mathrm{~kg}^{-1}}\\)
        \n= 1.0753
        \nEach molecule of CH2<\/sub>F COOH dissociate into 2 particles as
        \nCH2<\/sub>F COOH \\(\\rightleftharpoons\\) CH2<\/sub>F COO–<\/sup> + H+<\/sup>
        \n\"TS<\/p>\n

        Question 48.
        \n100g of liquid A(molar mass 140g mol-1<\/sup>) was dissolved in 1000g of liquid B(molar mass 180g mol-1<\/sup>). 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.
        \nAnswer:
        \nLet vapour pressure of pure A be \\(\\mathbf{p}_{\\mathrm{A}}^0\\)
        \n\\(p_{\\mathrm{B}}^0\\) = 500 torr ; nA<\/sub> = \\(\\frac{100}{140}\\) = 0.714 mol
        \nnB<\/sub> = \\(\\frac{1000}{180}\\) = 5.55 mol
        \nxA<\/sub> = \\(\\frac{0.714}{5.55+0.714}\\) = 0.114
        \nxB<\/sub> = 1 – 0.114 = 0.886
        \nptotal<\/sub> = \\(p_A^0 x+p_B^0 x_B\\)
        \n(or) 475(torr) = \\(p_A^0\\) \u00d7 0.114 + 500 \u00d7 0.886
        \n(or) \\(\\mathrm{p}_{\\mathrm{A}}^0\\) = 280.7 torr
        \nPA<\/sub> = 280.70 \u00d7 0.114 = 32 Torr<\/p>\n

        Question 49.
        \nDetermine the amount of CaCl2<\/sub> (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27\u00b0C.
        \nAnswer:
        \nWe know that \u03c0 = CRT
        \nor \u03c0 = \\(\\frac{\\mathrm{i} \\times \\mathrm{W}_{\\mathrm{B}} \\times \\mathrm{R} \\times \\mathrm{T}}{\\mathrm{M}_{\\mathrm{B}} \\times \\mathrm{V}_{\\mathrm{L}}}\\)
        \nor WB<\/sub> = \\(\\frac{\\pi \\times M_B \\times V_L}{i \\times R \\times T}\\)
        \nvan\u2019t Hoff factor i = 2.47
        \nOsmotic pressure p = 0.75 atm.
        \ntemperature T = 273 + 27 = 300 K.
        \nMol. wt of CaCl2<\/sub>, MB<\/sub> = 111
        \nVolume of solution = 2.5 L
        \nConstant R = 0.0821 L atm mol-1<\/sup> K-1<\/sup>
        \nWB<\/sub> = \\(\\frac{0.75 \\times 111 \\times 2.5}{2.7 \\times 0.0821 \\times 300}\\) = 3.40 g
        \nMoles of CaCl2<\/sub> = \\(\\frac{3.4}{111}\\) = 0.03 mol.<\/p>\n

        Question 50.
        \nDetermine the osmotic pressure of a solution prepared by dissolving 25 mg of K2<\/sub>SO4<\/sub> in two litre of water at 25\u00b0C assuming that it is completely disassociated.
        \nAnswer:
        \nK2<\/sub>SO4<\/sub> is strong electrolyte, ionises completely
        \nK2<\/sub>SO4<\/sub> \u2192 2K+<\/sup> + \\(\\mathrm{SO}_4^{2-}\\)
        \nTherefore i = 3
        \nVolume of the solution = 2L
        \nMass of K2<\/sub>SO4<\/sub> (WB<\/sub>) = 25 mg = 0025 g
        \nMolar mass of K2<\/sub>SO4<\/sub> (MB<\/sub>) = 174 g mol-1<\/sup>
        \n\u03c0 = iCRT = i \u00d7 \\(\\frac{3 \\times 0.025 \\times 0.0821 \\times 298}{174 \\times 2}\\)
        \n= 5.27 \u00d7 10-3<\/sup> atm<\/p>\n

        Question 51.
        \nBenzene 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.
        \nAnswer:
        \nIf yb<\/sub> is the mole fraction of benzene in vapour form
        \n\"TS
        \n\"TS<\/p>\n

        Intext Questions – Answers<\/span><\/p>\n

        Question 1.
        \nCalculate the mass percentage of benzene (C6<\/sub>H6<\/sub>) and carbon tetrachloride (CCl4<\/sub>) if 22g of benzene is dissolved in 122g of carbon tetrachloride.
        \nAnswer:
        \nWeight of Benzene, w = 22 g
        \nWeight of carbon tetrachloride, W = 122 g
        \nMass percentage of benzene = \\(\\frac{w \\times 100}{W+w}\\)
        \n= \\(\\frac{22 \\times 100}{22+122}\\) = 15.27%
        \nMass percentage of carbon tetrachloride
        \n= \\(\\frac{W \\times 100}{W+w}\\) = \\(\\frac{122 \\times 100}{22+122}\\) = 84.73%<\/p>\n

        Question 2.
        \nCalculate the mole fraction of benzene in solution containing 30% by mass of it in carbon tetrachloride.
        \nAnswer:
        \nMass of benzene, w = 30 g
        \nMol. wt of benzene = 78 g mol-1<\/sup>
        \nMoles of benzene = \\(\\frac{30}{78}\\) = 0.385
        \nMass of CCl4<\/sub> = 70 g
        \nMol. wt. of CCl4<\/sub> = 154 g mol-1<\/sup>
        \nMoles of CCl4<\/sub> = \\(\\frac{70(\\mathrm{~g})}{154(\\mathrm{~g})}\\) = 0.454
        \nMole fraction of benzene = \\(\\frac{0.385}{0.385+0.454}\\)
        \nMole fraction of CCl4<\/sub> = (1 – XBenzene<\/sub>)
        \n= 1 – 0.459 = 0.541<\/p>\n

        Question 3.
        \nCalculate the molarity of each of the following solutions
        \n(a) 30 g of CO(NO3<\/sub>)2<\/sub>. 6H2<\/sub>O in 4.3L of solution
        \n(b) 30 ml of 0.5 M H2<\/sub>SO4<\/sub> diluted to 500 ml.
        \nAnswer:
        \na) Mass of CO (NO3<\/sub>)2<\/sub>. 6H2<\/sub>0 = 30 g
        \nMolar mass of CO(NO3<\/sub>)2<\/sub>. 6H2<\/sub>O
        \n= 297 g mol-1<\/sup>.
        \nVolume of solution = 4.3 L
        \nMolarity M = \\(\\frac{30(\\mathrm{~g})}{297 \\mathrm{gmol}^{-1} \\times 4.3 \\mathrm{~L}}\\)
        \n= 0.023 mol L-1<\/sup><\/p>\n

        b) Initial volume V1<\/sub> = 30 ml
        \nMolarity M1<\/sub> = 0.5 mol L-1<\/sup>
        \nFinal volume V2<\/sub> = 500 ml
        \nFinal molarity M2<\/sub> =?
        \nM1<\/sub>V1<\/sub> = M2<\/sub>V2<\/sub> (or) M2<\/sub> = \\(\\frac{M_1 V_1}{V_2}\\)
        \n= \\(\\frac{0.5 \\times 30}{500}\\) = 0.03 mol L-1<\/sup><\/p>\n

        Question 4.
        \nCalculate the mass of urea (NH2<\/sub>CONH2<\/sub>) required in makIng 2.5 kg of 0.25 molal aqueous solution.
        \nAnswer:
        \nThe mass of urea required = WB<\/sub>
        \nMolar mass of urea MB<\/sub> = 60
        \nMass of solution WA<\/sub> + WB<\/sub> = 2.5 kg = 2500 gm.
        \nMass of solvent WA<\/sub> = 2500 – WB<\/sub>
        \nMolarity m = 0.25 mol kg-1<\/sup>
        \nm = \\(\\frac{\\mathrm{W}_{\\mathrm{B}} \\times 1000}{\\mathrm{M}_{\\mathrm{A}} \\times \\mathrm{W}_{\\mathrm{B}}}\\)
        \n0.25 = \\(\\frac{\\mathrm{W}_{\\mathrm{B}} \\times 1000}{6.0 \\times\\left(2500-\\mathrm{W}_{\\mathrm{B}}\\right)}\\)
        \nSolving we get WB<\/sub> = 37.5 g.<\/p>\n

        \"TS<\/p>\n

        Question 5.
        \nCalculate
        \n(a) molality
        \n(b) molarity
        \n(c) mole fraction of KI if the density of 20% (mass\/ mass) aqueous KI is 1.202 g mL-1<\/sup>
        \nAnswer:
        \nMass of KI(WB<\/sub>) = 20 g
        \nMolar mass of KI (MB<\/sub>) = 166 g mol-1<\/sup>
        \nDensity of solution (d) = 1.202 g mL-1<\/sup>
        \nMass of solution (WA<\/sub> + WB<\/sub>) = 100 g
        \nVolume of solution (V) = \\(\\frac{\\text { Mass }}{\\text { density }}\\)
        \n= \\(\\frac{100}{1.202 \\mathrm{gmL}^{-1}}\\) = 83.2 ml
        \nMass of solvent WA<\/sub> = 100 – 20 = 80 g.<\/p>\n

        a) Molality (m) = \\(\\frac{W_B \\times 1000}{M_B \\times W_A}\\)
        \n= \\(\\frac{20 \\mathrm{~g} \\times 1000 \\mathrm{~g} \\mathrm{~kg}^{-1}}{166 \\mathrm{~g} \\mathrm{~mol}^{-1} \\times 80 \\mathrm{~g}}\\)
        \n= 1.51 mol kg-1<\/sup><\/p>\n

        b) Molarity (M) = \\(\\frac{\\mathrm{W}_{\\mathrm{B}} \\times 1000}{\\mathrm{M}_{\\mathrm{B}} \\times \\mathrm{V}}\\)
        \n= \\(\\frac{20 \\mathrm{~g} \\times 1000 \\mathrm{~mL}}{166 \\mathrm{~g} \\mathrm{~mol}^{-1} \\times 83.2 \\mathrm{ml}}\\)
        \n= 1.45 mol L-1<\/sup><\/p>\n

        c)
        \nMoles of KI(nB<\/sub>) = \\(\\frac{20 \\mathrm{~g}}{166 \\mathrm{~g} \\mathrm{~mol}^{-1}}\\) = 1.51 mol
        \nMoles of water (nA<\/sub>) = \\(\\frac{1000 \\mathrm{~g}}{18 \\mathrm{~g} \\mathrm{~mol}^{-1}}\\) = 55.5
        \n\u2234 Mole fraction of KI (xB<\/sub>) = \\(\\frac{1.51}{1.51+55.5}\\)
        \n= 0.026<\/p>\n

        Question 6.
        \nH2<\/sub>S, a toxic gas with rotten egg like smell, is used for qualitative analysis. If the solubility of H2<\/sub>S in water at STP is 0.195 m, calculate Henry\u2019s law constant.
        \nAnswer:
        \nAccording to Henry\u2019s law
        \nKH<\/sub> . x = p or KH<\/sub> = \\(\\frac{p}{x}\\)
        \nP = 1 bar, solubility = 0.195 mol kg-1<\/sup>
        \nMoles of H2<\/sub>S (nB<\/sub>) = 0.195
        \nMoles of water (nA<\/sub>) = \\(\\frac{1000 \\mathrm{~g}}{18 \\mathrm{~g} \\mathrm{~mol}^{-1}}\\) = 55.5
        \n\u2234 Mole fraction of H2<\/sub>S (x)
        \n= \\(\\frac{0.195}{0.195+55.5}\\) = 3.5 \u00d7 10-3<\/sup>
        \nKH<\/sub> = \\(\\frac{1(\\text { bar) }}{3.5 \\times 10^{-3}}\\) = 285.6 bar<\/p>\n

        Question 7.
        \nHenry\u2019s law constant of CO2<\/sub> in water is 1.67 \u00d7 108<\/sup> Pa at 298K. Calculate the quantity of CO2<\/sub> in 500 mL of soda water when packed under 25 atm CO2<\/sub> pressure at 298 K.
        \nAnswer:
        \nPressure of CO2<\/sub> (p) = 2.5 atm
        \n= 2.5 atm \u00d7 101325\\(\\frac{(\\mathrm{Pa})}{\\mathrm{atm}}\\)
        \n= 253312.5 Pa
        \nAccording to Henrys law
        \n\"TS
        \nMoles of CO2<\/sub> present in 500 ml
        \n= \\(\\frac{0.0844}{2}\\) mol = 0.0422 mol.
        \nAmount of CO2<\/sub> present in 500 ml
        \n= 0.0422 mol \u00d7 440 g mol-1<\/sup> = 1.86 g.<\/p>\n

        Question 8.
        \nThe 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.
        \nAnswer:
        \nTotal vapour pressure PT<\/sub> = 600 mm Hg
        \nLet Mole fraction of A is XA<\/sub> and that of B is XB<\/sub>
        \nVapour pressure of pure A(\\(\\mathrm{P}_{\\mathrm{A}}^0\\)) = 450 mm Hg
        \nVapour pressure of pure B(\\(\\mathrm{P}_{\\mathrm{B}}^0\\)) =700 mm Hg
        \nNow PTotal<\/sub> = \\(\\mathrm{P}_{\\mathrm{A}}^0 \\mathrm{x}_{\\mathrm{A}}\\) + \\(\\mathrm{P}_{\\mathrm{B}}^0 \\mathrm{x}_{\\mathrm{B}}\\)
        \n\u2234 600 = 450 xA<\/sub> + 700 xB<\/sub>
        \n= 450xA<\/sub> + 700(1 – xA<\/sub>)
        \n= 700 – 250 xA<\/sub> or xA<\/sub> = 0.4
        \nxB<\/sub> = 1 – xA<\/sub> = 1 – 0.4 = 0.6
        \nMole fraction of A in vapour phase
        \nYA<\/sub> = \\(\\frac{x_A P_A^0}{P_{\\text {Total }}^0}\\) = \\(\\frac{0.4 \\times 450}{600}\\) = 0.3
        \nMole fraction of B in vapour phase
        \nYB<\/sub> = \\(\\frac{\\mathrm{x}_{\\mathrm{B}} \\mathrm{P}_{\\mathrm{B}}^0}{\\mathrm{P}_{\\text {Total }}}\\) = \\(\\frac{0.6 \\times 700}{600}\\) = 0.7<\/p>\n

        Question 9.
        \nVapour pressure of pure water at 298K is 23.8 mm Hg. 50 of urea (NH2<\/sub> CO NH2<\/sub>) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
        \nAnswer:
        \nMass of water = 850 g
        \nMoles of water = \\(\\frac{850}{18}\\) = 47.22 mol
        \nMass of urea = 50 g
        \nMol mass of urea = 60
        \nMoles of urea = \\(\\frac{50}{60}\\) = 0.83
        \nSince the solution is not dilute the formula to be used is
        \n\"TS<\/p>\n

        Question 10.
        \nBoiling point of water at 750 min Hg is 99.63\u00b0C. How much sucrose is to be added to 500g of water such that it boils at 100\u00b0C.
        \nAnswer:
        \nLet mass of sucrose required = WB<\/sub> g
        \nMass of solvent WA<\/sub> = 500 g
        \nMolar mass of sucrose (C12<\/sub>H22<\/sub>O11<\/sub>) = 342
        \nElevation in boiling point \u0394Tb<\/sub> = 100 – 99.63
        \n= 0.37\u00b0C
        \n\u0394Tb<\/sub> = \\(\\frac{\\mathrm{K}_{\\mathrm{b}} \\times \\mathrm{W}_{\\mathrm{B}} \\times 1000}{\\mathrm{~W}_{\\mathrm{A}} \\times \\mathrm{M}_{\\mathrm{B}}}\\)
        \n(or) WB<\/sub> = \\(\\frac{\\Delta \\mathrm{T}_{\\mathrm{b}} \\times \\mathrm{W}_{\\mathrm{A}} \\times \\mathrm{M}_{\\mathrm{B}}}{\\mathrm{K}_{\\mathrm{b}} \\times 1000}\\)
        \n= \\(\\frac{0.37 \\times 500 \\times 342}{0.52 \\times 1000}\\) = 121.<\/p>\n

        Question 11.
        \nCalculate the mass of ascorbic acid (vitamin C, C6<\/sub>H8<\/sub>O6<\/sub>) to be dissolved in 75g of
        \nacetic acid to lower its melting point by 1.5\u00b0C. Kf<\/sub> = 3.9 K kg mol-1<\/sup>.
        \nAnswer:
        \nLet the mass of ascorbic acid required = WB<\/sub>g
        \nMolar mass of ascorbic acid (C6<\/sub>H8<\/sub>O6<\/sub>) MB<\/sub> = 176 g mol-1<\/sup>
        \nMass of solvent (WA<\/sub>) = 75 g
        \nDepression in melting point \u0394Tf<\/sub> = 1.5\u00b0C
        \n\"TS<\/p>\n

        Question 12.
        \nCalculate 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\u00b0C.
        \nAnswer:
        \nMass of polymer (WB<\/sub>) = 1.0 g
        \nMolar mass of polymer MB<\/sub> = 1,85,000 g mol-1<\/sup>
        \nT = 37 + 273.15 = 310.15K
        \n\u03c0 = \\(\\frac{W_B R_T}{M_B V_{(L)}}\\)
        \n= \\(\\frac{1 \\times 0.082 \\times 310.15}{185000 \\times 450 \\times 10^{-3}}\\)
        \n= 3.05 \u00d7 10-4<\/sup> atm
        \n= 3.05 \u00d7 10-4<\/sup> \u00d7 1.01 \u00d7 105<\/sup>Pa = 30.9 Pa<\/p>\n

        Problems<\/span><\/p>\n

        Question 1.
        \nCalculate the mole fraction of ethylene glycol (C2<\/sub>H6<\/sub>O2<\/sub>) in a solution containing 20% of C2<\/sub>H6<\/sub>O2<\/sub> by mass. (TS 15)
        \nAnswer:
        \nAssume 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.
        \nMolar mass of C2<\/sub>H6<\/sub>O2<\/sub> = 12 \u00d7 2 + 1 \u00d7 6 + 16 \u00d7 2 = 62 g mol-1<\/sup>.
        \n\"TS<\/p>\n

        Question 2.
        \nCalculate the molarity of a solution containing 5 g of NaOH in 450 mL solution. (TS Mar. 19; AP & TS 15)
        \nAnswer:
        \nMoles of NaOH = \\(\\frac{5 \\mathrm{~g}}{40 \\mathrm{~g} \\mathrm{~mol}^{-1}}\\) = 0.125 mol
        \nVolume of the solution in litres = 450 mL \/1000 mL L-1<\/sup>
        \nUsing equation (2.8)
        \nMolarity = \\(\\frac{0.125 \\mathrm{~mol} \\times 1000 \\mathrm{~mL} \\mathrm{~L}^{-1}}{450 \\mathrm{~mL}}\\)
        \n= 0.278 M = 0.278 mol L-1<\/sup>
        \n= 0.278 mol dm-3<\/sup><\/p>\n

        Question 3.
        \nCalculate molality of 2.5 g of ethanoic acid (CH3<\/sub>COOH) in 75 g of benzene. (TS 15)
        \nAnswer:
        \nMolar mass of C2<\/sub>H4<\/sub>O2<\/sub> = 12 \u00d7 2 + 1 \u00d7 4 + 16 \u00d7 2 = 60 g mol-1<\/sup>.
        \nMoles mass of C2<\/sub>H4<\/sub>O2<\/sub> = \\(\\frac{2.5 \\mathrm{~g}}{60 \\mathrm{~g} \\mathrm{~mol}^{-1}}\\)
        \n= 0.0417 mol
        \nMass of benzene in kg = 75 g \/ 1000 g kg-1<\/sup>
        \n= 75 \u00d7 10-3<\/sup> kg
        \nMolality of C2<\/sub>H4<\/sub>O2<\/sub> = \\(\\frac{\\text { moles of } \\mathrm{C}_2 \\mathrm{H}_4 \\mathrm{O}_2}{\\text { kg of benzene }}\\)
        \n= \\(\\frac{0.0417 \\mathrm{~mol} \\times 1000 \\mathrm{~g} \\mathrm{~kg}^{-1}}{75 \\mathrm{~g}}\\)
        \n= 0.556 mol kg-1<\/sup><\/p>\n

        \"TS<\/p>\n

        Question 4.
        \nIf N2<\/sub> gas is bubbled through water at 293K, how many mililmoles of N2<\/sub> gas would dissolve in 1 litre of water? Assume that N2<\/sub> exerts a partial pressure of 0.987 bar. Given that Henry\u2019s law constant for N2<\/sub> at 293 K is 76.48 kbar.
        \nAnswer:
        \nThe 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\u2019s law. Thus:
        \n\"TS
        \nAs 1 litre of water co\u00f1tains 55.5 mol of it, therefore If n represents number of moles of N2<\/sub> in solution.
        \nx(nitrogen) = \\(\\frac{n \\mathrm{~mol}}{\\mathrm{n} \\mathrm{mol}+55.5 \\mathrm{~mol}}\\)
        \n= \\(\\frac{\\mathrm{n}}{55.5}\\) = 1.29 \u00d7 10-5<\/sup>
        \n(n in denominator is neglected as it is <<55.5)
        \nThus n = 1.29 \u00d7 10-5<\/sup> \u00d7 55.5 mol
        \n= 7.16 \u00d7 10-4<\/sup> mol-1<\/sup>
        \n= \\(\\frac{7.16 \\times 10^{-4} \\mathrm{~mol} \\times 1000 \\mathrm{~m} \\mathrm{~mol}}{1 \\mathrm{~mol}}\\)
        \n= 0.716 m mol<\/p>\n

        Question 5.
        \nVapour pressure of chloroform (CHCl3<\/sub>) and dichloromethane (CH2<\/sub>Cl2<\/sub>) at 298 K are 200mm Hg and 415 mm Hg respectively.
        \n(i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl3<\/sub> and 40 g of CH2<\/sub>Cl2<\/sub> at 298 K and,
        \n(ii) mole fractions of each component in vapour phase.
        \nAnswer:
        \ni) Molar mass of CH2<\/sub>Cl2<\/sub> = 12 \u00d7 1 + 1 \u00d7 2 + 35.5 \u00d7 2 = 85 g mol-1<\/sup>
        \nMolar mass of CHCl3<\/sub> = 12 \u00d7 1 + 1 \u00d7 1 + 35.5 \u00d7 3 = 119.5 g mol-1<\/sup>
        \nMoles of CH2<\/sub>Cl2<\/sub> = \\(\\frac{40 \\mathrm{~g}}{85 \\mathrm{~g} \\mathrm{~mol}^{-1}}\\) = 0.47 mol
        \nMoles of CHCl3<\/sub> = \\(\\frac{25.5 \\mathrm{~g}}{119.5 \\mathrm{~g} \\mathrm{~mol}^{-1}}\\)
        \n= 0.213 mol
        \nTotal number of moles = 0.47 + 0.213
        \n= 0.683 mol
        \n\"TS
        \nUsing equation (2.16),
        \nPtotal<\/sub> = \\(\\mathrm{p}_1^0\\) + (\\(\\mathrm{p}_2^0\\) + \\(\\mathrm{p}_1^0\\))x2<\/sub>
        \n= 200 + (415 – 200) \u00d7 0.688
        \n= 200 + 147.9 = 347.9 mm Hg<\/p>\n

        (ii) Using the relation (2.19) y1<\/sub> = P1<\/sub> \/ ptotal<\/sub>, we can calculate the mole fraction of the components in gas phase (y1<\/sub>).
        \n\\(\\mathrm{p}_{\\mathrm{CH}_2} \\mathrm{Cl}_2\\) = 0.688 \u00d7 415 mm Hg = 285.5 mm Hg
        \n\\(\\mathbf{p}_{\\mathrm{CHCl}_3}\\) = 0.312 \u00d7 200 mm Hg = 62.4 mm Hg
        \n\\(\\mathbf{y}_{\\mathrm{CH}_2 \\mathrm{Cl}_2}\\) = 285.5 mm Hg \/ 347.9 mm Hg = 0.82
        \n\\(\\mathbf{y}_{\\mathrm{CHCl} 3}\\) = 62.4 mm Hg\/ 347.9 mm Hg = 0.18<\/p>\n

        Note : Since, CH2<\/sub>Cl2<\/sub> is a more volatile component than CHCl3<\/sub>, [\\(\\mathrm{p}_{\\mathrm{CH}_2}^0 \\mathrm{Cl}_2\\) = 415 mm Hg and \\(\\mathbf{p}_{\\mathrm{CHC} l_3}^0\\) = 200 mm Hg] and the vapour phase is also richer in CH2<\/sub>Cl2<\/sub> [\\(\\mathbf{y}_{\\mathrm{CH}_2} \\mathrm{Cl}_2\\) = 0.82 and \\(\\mathbf{y}_{\\mathrm{CHCl}_3} \\)= 0.18], it may thus be concluded that at equilibrium, vapour phase will be always rich in the component which is more volatile.<\/p>\n

        Question 6.
        \nThe vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non – volatile, non – electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol-1<\/sup>). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance ? (Mar. 2018, 16 AP)
        \nAnswer:
        \nThe various quantities known to us are as follows :
        \n\\(\\mathrm{p}_1^0\\) = 0.850 bar ; p = 0.845 bar
        \nM1<\/sub> = 78 g mol-1<\/sup>; w2<\/sub> = 0.5 g; w1<\/sub> = 39 g
        \nSubstituting these values in equation (2.28), we get
        \n\\(\\frac{0.850 \\mathrm{bar}-0.845 \\mathrm{bar}}{0.850 \\mathrm{bar}}\\) = \\(\\frac{0.5 \\mathrm{~g} \\times 78 \\mathrm{~g} \\mathrm{~mol}^{-1}}{\\mathrm{M}_2 \\times 39 \\mathrm{~g}}\\)
        \nTherefore, M2<\/sub> = 170 g mol-1<\/sup><\/p>\n

        Question 7.
        \n18 g of glucose, C6<\/sub>H12<\/sub>O6<\/sub>, is dissolved in 1 kg of water In a saucepan. At what temperature will water boll at 1.013 bar? Kb<\/sub> for water is 0.52 K kg mol-1<\/sup>.
        \nAnswer:
        \nMoles of glucose = 18 g \/ 180 g mol-1<\/sup>
        \n= 0.1 mol
        \nNumber of kilograms of solvent = 1 kg
        \nThus molality of glucose solution
        \n= 0.1 mol kg-1<\/sup>
        \nFor water, change in boiling point
        \n\u0394Tb<\/sub> = Kb<\/sub> \u00d7 m = 0.52 K kg mol-1<\/sup> \u00d7 0.1 mol kg-1<\/sup> = 0.052 K<\/p>\n

        Since water boils at 373.15 Kat 1.013 bar pressure, therefore, the boiling point of solution will be 373.15 + 0.052 = 373.202 K.<\/p>\n

        Question 8.
        \nThe boiling point of benzene is 353.23 K. When 1.80 g of a non – volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. Kb for benzene is 2.53 K kg mol-1<\/sup>.
        \nAnswer:
        \nThe elevation (\u0394Tb<\/sub>) in the boiling point = 354.11 K – 352.23 K = 0.88 K
        \nSubstituting these values in expression (2.33) we get
        \nM2<\/sub> = \\(\\frac{2.53 \\mathrm{~K} \\mathrm{~kg} \\mathrm{~mol}^{-1} \\times 1.8 \\mathrm{~g} \\times 1000 \\mathrm{~g} \\mathrm{~kg}^{-1}}{0.88 \\mathrm{~K} \\times 90 \\mathrm{~g}}\\)
        \n= 58 g mol-1<\/sup>
        \nTherefore, molar mass of the solute, M2<\/sub> = 58 g mol-1<\/sup><\/p>\n

        Question 9.
        \n45 g of ethylene glycol (C2<\/sub>H6<\/sub>O2<\/sub>) is mixed with 600 g of water. Calculate
        \n(a) the freezing point depression and
        \n(b) the freezing point of the solution.
        \nAnswer:
        \nDepression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol
        \nmoles of ethylene glycol mass of water in kilogram
        \n\"TS
        \nMoles of ethylene glycol = \\(\\frac{45 \\mathrm{~g}}{62 \\mathrm{~g} \\mathrm{~mol}^{-1}}\\)
        \n= 0.73 mol
        \nMass of water in kg = \\(\\frac{600 \\mathrm{~g}}{1000 \\mathrm{~g} \\mathrm{~kg}^{-1}}\\) = 0.6 kg
        \nHence molality of ethylene glycol = \\(\\frac{0.73 \\mathrm{~g}}{0.60 \\mathrm{~kg}}\\)
        \n= 1.2 mol kg-1<\/sup>
        \nTherefore freezing point depression.
        \n\u0394T1<\/sub> = 1.86K kg mol-1<\/sup> \u00d7 1.2 mol kg-1<\/sup> = 2.2K
        \nFreezing point of the aqueous solution
        \n= 273.15 K – 2.2 K = 270.95 K<\/p>\n

        Question 10.
        \n1.00 g of a non – electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing
        \npoint depression constant of benzene is 5.12 K mol-1<\/sup>. Find the molar mass of the solute.
        \nAnswer:
        \nSubstituting the values of various terms involved in equation (2.36) we get.
        \nM2<\/sub> = \\(\\frac{5.12 \\mathrm{~K} \\mathrm{~kg} \\mathrm{~mol}^{-1} \\times 1.00 \\mathrm{~g} \\times 1000 \\mathrm{~g} \\mathrm{~kg}^{-1}}{0.40 \\times 50 \\mathrm{~g}}\\)
        \n= 256 g mol-1<\/sup>
        \nThus, molar mass of the solute = 256 g mol-1<\/sup><\/p>\n

        Question 11.
        \n200 cm3<\/sup> of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 \u00d7 10-3<\/sup> bar. Calculate the molar mass of the protein.
        \nAnswer:
        \nThe various quantities known to us are as follows : \u03c0 = 2.57 \u00d7 10-3<\/sup> bar,
        \nV = 200 cm3<\/sup> = 0.200 litre
        \nT = 300 K
        \nR = 0.083 L bar mol-1<\/sup>K-1<\/sup>
        \nSubstituting these values in equation (2.42) we get
        \nM2<\/sub> = \\(\\frac{1.26 \\mathrm{~g} \\times 0.083 \\mathrm{~L} \\mathrm{bar} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1} \\times 300 \\mathrm{~K}}{2.57 \\times 10^{-3} \\mathrm{bar} \\times 0.200 \\mathrm{~L}}\\)
        \n= 61.022 g mol-1<\/sup><\/p>\n

        Question 12.
        \n2 g of benzoic acid (C6<\/sub>H5<\/sub>COOH) dissolved in 25 g of benzene shows a depression in freezing point equal 101.62K. Molal depression constant for benzene is 4.9 K kg mol-1<\/sup>. What is the percentage association of acid if it forms dimer in solution?
        \nAnswer:
        \nThe given quantities are : w2<\/sub> = 2 g;
        \nKf<\/sub> = 4.9 K kg mol-1<\/sup> ; w1<\/sub> = 25 g,
        \n\u0394Tf<\/sub> = 1.62 K
        \nSubstituting these values in equation (2.36) we get:
        \nM2<\/sub> = \\(\\frac{4.9 \\mathrm{~K} \\mathrm{~kg} \\mathrm{~mol}^{-1} \\times 2 \\mathrm{~g} \\times 1000 \\mathrm{~g} \\mathrm{~kg}^{-1}}{25 \\mathrm{~g} \\times 1.62 \\mathrm{~K}}\\)
        \n= 241.98 g mol-1<\/sup>
        \nThus, experimental molar mass of benzoic acid in benzene is = 241.98 g mol-1<\/sup>
        \nNow consider the following equilibrium for the acid:
        \n2C6<\/sub>H5<\/sub>COOH \\(\\rightleftharpoons\\) (C6<\/sub>H5<\/sub>COOH)2<\/sub>
        \nIf x represents the degree of association of the solute then we would have (1 – x) mol of benzoic acid left in unassociated form and correspondingly \\(\\frac{x}{2}\\) as associated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is :
        \n1 – x + \\(\\frac{x}{2}\\) = 1 – \\(\\frac{x}{2}\\)
        \nThus, total number of moles of particles at equilibrium equals van\u2019t Hoff factor i.
        \n\"TS
        \nTherefore, degree of association of benzoic acid in benzene is 99.2%.<\/p>\n

        \"TS<\/p>\n

        Question 13.
        \n0.6 mL of acetic acid (CH3<\/sub>COOH), having density 1.06 g mL-1<\/sup>, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205\u00b0C. Calculate the van\u2019t Hoff factor and the dissociation constant of acid.
        \nAnswer:
        \nNumber of moles of acetic acid
        \n= \\(\\frac{0.6 \\mathrm{~mL} \\times 1.06 \\mathrm{~g} \\mathrm{~mL}^{-1}}{60 \\mathrm{~g} \\mathrm{~mol}^{-1}}\\) = 0.0106 mol = n
        \nMolarity = \\(\\frac{0.0106 \\mathrm{~mol}}{1000 \\mathrm{~mL} \\times 1 \\mathrm{~g} \\mathrm{~mL}^{-1}}\\)
        \n= 0.0106 mol kg-1<\/sup>
        \nUsing equation (2.35)
        \n\u2206Tf<\/sub> = 1.86 K kg mol-1<\/sup> \u00d7 0.0106 mol kg-1<\/sup> = 0.0197K
        \nvan\u2019t Hoff factor
        \n\"TS
        \nAcetic acid is a weak electrolyte and will dissociate into two ions : acetate and hydrogen ions per molecule of acetic acid. If x is the degree of dissociation of acetic acid, then we would have n (1 – x) moles of undisso-ciated acetic acid, nx moles of CH3<\/sup>COO–<\/sup> and nx moles of H+<\/sup> ions.
        \n\"TS
        \nThus total moles of particles are :
        \nn(1 – x + x + x) = n(1 + x)
        \ni = \\(\\frac{n(1+x)}{n}\\) = 1 + x = 1.041
        \nThus degree of dissociation of acetic acid = x = 1.041 – 1.000 = 0.041
        \nThen [CH3<\/sub>COOH] = n(1 – x)
        \n= 0.0106 ( 1 – 0.041),
        \n[CH3<\/sub>COO–<\/sup>] = nx = 0.0106 \u00d7 0.041, [H+<\/sup>]
        \n= nx = 0.0106 \u00d7 0.041.
        \n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

        Telangana TSBIE\u00a0TS 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. 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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: ... 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