Solution and Colligative Properties Exercise # 5

SOLUTION AND COLLIGATIVE PROPERTIES Exercise # 5 1. Part # I [Previous Year Questions 6.02 × 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is : (A) 0.001 M (B) 0.01 M (C) 0.02 M (D) 0.1 M. 2. Which one of the following aqueous solutions will exhibit highest boiling point ? (A) 0.01 M Na2SO4 (B) 0.01 M KNO3 (C) 0.015 M urea (D) 0.015 M glucose 3. Equimolar solutions in the same solvent have : (A) same boiling point but different freezing point (C) same boiling and same freezing points (B) same freezing point but different boiling point (D) different boiling and freezing points 4. Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 mL of 1.2 M second solution. What is the molarity of the final mixture ? (A) 1.20 M (B) 1.50 M (C) 1.344 M (D) 2.70 M 5. Benzene and toluene form nearly ideal solutions. At 20ºC, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20 ºC for a solution containing 78 g of benzene and 46 g of toluene in torr is : (A) 50 (B) 25 (C) 37.5 (D) 53.5 6. If  is the degree of dissociation of Na2SO4, the vant Hoff’s factor (i) used for calculating the molecular mass is : (A) 1 +  (B) 1 –  (C) 1 + 2 (D) 1 – 2. 7. Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is (A) 3.28 mol Kg–1 (B) 2.28 mol Kg–1 (C) 0.44 mol Kg–1 (D) 1.14 mol Kg–1 8. A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be (A) 700 (B) 360 (C) 350 (D) 300 9. A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60g mol–1) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm–3, molar mass of the substance will be (A) 105.0 g mol–1 (B) 210.0 g mol–1 (C) 90.0 g mol–1 (D) 15.0 g mol–1 10. The vapour pressure of water at 20º C is 17.5 mm Hg. If 18g of glucose (C6H12O6) is added to 178.2 g of water at 20° C, the vapour pressure of the resulting solution will be (A) 15.750 mm Hg (B) 16.500 mm Hg (C) 17.325 mm Hg (D) 17.675 mm Hg 11. At 80º C , the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 mm Hg. If a mixture solution of 'A' and 'B' boils at 80º C and 1 atm pressure, the amount of 'A' in the mixture is (1atm = 760 mm Hg) (A) 34 mol percent 12. (B) 48 mol percent (C) 50 mol percent (D) 52 mol percent A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statement is correct regarding the behaviour of the solution ? (A) The solution is non-ideal, showing +ve deviation from Raoult’s Law. (B) The solution in non-ideal, showing –ve deviation from Raoult’s Law. (C) n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s Law. (D) The solution formed is an ideal solution. 13. Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively : (A) 300 and 400 14. 15. 16. 17. 18. 19. 18g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is: (A) 76.0 (B) 752.4 (C) 759.0 (D) 7.6 The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : (Kf for benzene = 5.12 K kg mol–1) (A) 64.6 % (B) 80.4 % (C) 74.6 % (D) 94.6 % For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point ? (B) [Co(H2O)4Cl2]Cl.2H2O (C) [Co(H2O)3Cl3].3H2O (D) [Co(H2O)6]Cl3 An aqueous solution contains an unknown concentration of Ba . When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 × 10–10. What is the original concentration of Ba2+ ? 2+ Part # II 2. (D) 200 and 300 (A) 72.0 kPa (B) 36.1 kPa (C) 96.2 kPa (D) 144.5 kPa Kf for water is 1.86 K kg mol–1. If your automobile radiator holds 1.0 kg of water, how may grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8ºC ? (A) 72 g (B) 93 g (C) 39 g (D) 27 g (A) 2 × 10–9 M 1. (C) 500 and 600 If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (Tf), when 0.01 mole of sodium sulphate is dissolved in 1 kg of water, is (Kf = 1.86 K kg mol–1) (A) 0.0372 K (B) 0.0558 K (C) 0.0744 K (D) 0.0186 K On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane - 100 g mol–1 and of octane = 114 g mol–1) (A) [Co(H2O)5Cl]Cl2.H2O 20. (B) 400 and 600 (B) 1.1 × 10–9 M (C) 1.0 × 10–10 M (D) 5 × 10–9 M [Previous Year Questions] A 0.004 M solution of Na2SO4 is isotonic with 0.010 M solution of glucose at same temperature. The apparent percentage dissociation of Na2SO4 is : (A) 25% (B) 50% (C) 75% (D) 85% 1.22 g of benzoic acid is dissolved in 100 g of acetone and 100 g of benzene separately. Boiling point of the solution in acetone increases by 0.17ºC, while that of, in the benzene increases by 0.13ºC ; Kb for acetone and benzene is 1.7 K kg mol–1 and 2.6 K kg mol–1. Find molecular weight of benzoic acid in two cases and justify your answer. 3. The elevation in boiling point of a solution of 13.44 g of CuCl2 in 1kg of water using the following information, will be (Molecular weight of CuCl2 = 134.4 and Kb = 0.52 K molal–1) : (A) 0.16 (B) 0.05 (C) 0.1 (D) 0.2 4. When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol–1), a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is : (A) 0.5 (B) 1 (C) 2 (D) 3 SOLUTION AND COLLIGATIVE PROPERTIES Comprehension Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 Given : Freezing point depression constant of water (Kfwater) = 1.86 K kg mol–1 Freezing point depression constant of ethanol (Kfethanol) = 2.0 K kg mol–1 Boiling point elevation constant of water (Kbwater) = 0.52 K kg mol–1 Boiling point elevation constant of ethanol (Kbethanol) = 1.2 K kg mol–1 Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol–1 Molecular weight of ethanol = 46 g mol–1 In answering the following questions, consider the solution to be ideal dilute solutions and solutes to be nonvolatile and non-dissociative. 5. The freezing point of the solution M is (A) 268.7 K 6. (B) 36.0 mm Hg (C) 29.5 mm Hg (D) 28.8 mm Hg (B) 376.2 K (C) 375.5 K (D) 354.7 K The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water of 298 K and 5 atm pressure is : (A) 4 × 10–4 9. (D) 150.9 K Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is (A) 380.4 K 8. (C) 234.2 K The vapour pressure of the solution M is (A) 39.3 mm Hg 7. (B) 268.5 K (B) 4.0 × 10–5 (C) 5.0 × 10–4 (D) 4.0 × 10–6 The freezing point (in ºC) of a solution containing 0.1 g of K3[Fe(CN)6] (Mol. Wt. 329) in 100 g of water (Kf = 1.86 K kg mol–1) is : (A) – 2.3 × 10–2 (B) – 5.7 × 10–2 (C) – 5.7 × 10–3 (D) – 1.2 × 10–2 10. For a dilute solution containing 2.5 g of a non- volatile non- electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol–1) (A) 724 11. (B) 740 (C) 736 (D) 718 The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density  MWsolute  of this solution at 298 K is 2.0 g cm–3. The ratio of the molecular weights of the solute and solvent,   ,is  MWsolvent  12. For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here xL and xM represent mole fractions of L and M, respectively, in the solution. The correct statement(s) applicable to this system is(are) Z PL 1 XM 0 (A) The point Z represent vapour pressure of pure liquid M and Raoult's law is obeyed from xL = 0 to xL = 1 (B) Attractive intermolecular interactions between L–L in pure liquid L and M–M in pure liquid M are stronger than those between L-M when mixed in solution (C) The point Z represents vapour pressure of pure liquid M and Raoult's law is obeyed when xL0 (D) The point Z represents vapour pressure of pure liquid L and Raoult's law is obeyed when xL1 Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol of 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol–1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol–1] Among the following the option representing change in the freezing point is Ic e Wat er Wat er+ etha 1 (B) nol a r+e th Wate 270 273 T/K (C) Water Ic e r+ Wate 271 273 T/K nol 270 273 T/K et h an o l 1 (D) V.P./bar 1 Water Ic e V.P./bar (A) V.P./bar 1 V.P./bar 13. Ic e Wat er Wat er+ etha 271 273 T/K nol SOLUTION AND COLLIGATIVE PROPERTIES 14. Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions xA and xB, respectively, has vapour pressure of 22.5 Torr. The value of xA / xB in the new solution is_____. (given that the vapour pressure of pure liquid A is 20 Torr at temperature T) 15. The plot given below shows P–T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents. 3 1 2 4 760 1.Solvent X Pressure (mmHg) 2.Solution of NaCl in solvent X 3.Solvent Y 367 368 360 362 4.Solution of NaCl in solvent Y Temperature (K) On additionof equal number of moles of a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solvent S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is ___. EXERCISE - 5 : PART # I 1. B 14. B 2. A 15. A 3. C 16. B 4. C 17. B 5. A 18. D 6. C 19. C 7. B 20. B 8. C 9. B 10. C 11. C 12. A 13. B B 8. A 9. 10. A 11. 9 12. B, D PART # II 1. C 13. B 2. 122, 224 3. 14. 19 15. 0.05 A 4. A 5. D 6. B 7. A