Solution and Colligative Properties Exercise # 3

Exercise # 3 1. 2. [Matrix Match Type Questions] Part # I Column – I Column – II Assuming all the solutes are non volatile and all solutions are ideal and neglect the hydrolysis of cation and anion. (A) 10 ml 0.1 M NaOH aqueous solution is added to 10 ml 0.1 M HCl aqueous solution (B) 10 ml 0.1 M NaOH aqueous solution is added to 10 ml 0.1 M CH3COOH aqueous solution (C) 10 ml 0.1 M HCl aqueous solution is added to 10 ml 0.1 M NH3 aqueous solution (D) 10 ml 0.1 M HCl aqueous solution is added to 10 ml 0.1 M KOH aqueous solution (p) Osmotic pressure of solution increases Column I (A) Acetone + CHCl3 (B) Ethanol + Water (C) C2H5Br + C2H5I (D) Acetone + Benzene Column II (p) Smix. > 0 (q) Vmix. > 0 (r) Hmix. < 0 (s) Maximum boiling azeotropes (q) Vapour pressure of solution increases (r) Boiling point of solution increases (s) Freezing point of solution increases (t) Minimum boiling azeotropes 3. Column-I Column-II (Properties) (Affecting factors) (A) Relative lowering of vapour pressure (p) Directly proportional to van't Hoff factor, i (B) Elevation in boiling point (q) Directly proportional to molality (C) Freezing point (r) Directly proportional to molarity (D) Osmotic pressure (s) Indirectly proportional to lowering of vapour pressure Part # II [Comprehension Type Questions] Comprehension # 1 Addition of non–volatile solute to a solvent always increases the colligative properties such as osmotic pressure, P , Tb and Tf . All these colligative properties are directly proportional to molality if solutions are dilute. The increases in colligative properties on addition of non–volatile solute is due to increase in number of solute particles. 1. For different aqueous solutions of 0.1 N NaCI, 0.1 M urea , 0.1 N Na2SO4 and 0.1 N Na3PO4 solution at 270 C. The correct statements are : 1. The order of osmotic pressure is, NaCl > Na2SO4 > Na3PO4 > urea ΔTb 2.  = K × ST for urea solution b 3. Addition of salt on ice increases its melting point 4. Addition of salt on ice brings in melting of ice earlier (A) 2 , 3 ,4 (B) 2, 4 (C) 1 , 2 , 3 (D) 3 , 4 SOLUTION AND COLLIGATIVE PROPERTIES 2. 1 g mixture of glucose and urea present in 250 mL aqueous solution shows the osmotic pressure of 0.74 atm at 27 0 C. Assuming solution to be dilute , which are correct ? 1. Percentage of urea in mixture is 17.6. 2. Relative lowering in vapour pressure of this solution is 5.41 × 10–4. 3. The solution will boil at 100.0150 C , if Kb of water is 0.5 K molality–1. 4. If glucose is replaced by same amount of sucrose, the solution will show higher osmotic pressure at 27 0C. 5. If glucose is replaced by same amount of NaCl, the solution will show lower osmotic pressure at 270C. (A) 1 , 2 , 3 (B) 1 , 2 , 3 , 5 (C) 2 , 4 , 5 (D) 1 , 4 , 5 Comprehension # 2 Answer the questions (given below) which are based on the following diagram. P°A P total V.P. Consider some facts about the above phase diagram : P°B Vapour pressure diagram for real solutions of two liquids A and B that exhibit a positive deviation from Raoult’s law. The vapour pressure of both A and B are greater than predicted by Raoult’s law. The dashed lines represented the plots for ideal solutions. PA PB PB 0 XA 1 1 XB 0 1. A : This is observed when A...B attractions are greater than average of A...B and B...B attraction: B : Hmix = +ve, Vmix = +ve C : Boiling point is smaller than expected such that vaporisation is increased D : Mixture is called azeotropic mixture Select correct facts (A) A, B, C (B) B, C, D (C) A, C, D (D) A, B, C, D 2. Total vapour pressure of mixture of 1 mol of volatile component A (p°A = 100 mmHg) and 3 mol of volatile component B (p°B = 60 mmHg) is 75 mm. For such case : (A) There is positive deviation from Raoult’s law (B) Boiling point has been lowered (C) Force of attraction between A and B is smaller than that between A and A or between B and B. (D) All the above statements are correct. Comprehension # 3 Following passage explains effect of temperature on the vapour pressure of liquid. Answer the questions given at the end. Effect of temperature on Vapour pressure The quantity of heat required to evaporate a given liquid at constant temperature is defined as the heat of vaporisation. Variation of vapour pressure with temperature is given by Clausius-Clapeyron equation. loge P = – ΔH vap RT + log e A A liquid is said to be at its boiling temperature if its vapour pressure is equal to external pressure. Therefore, the boiling point of water in particular and of liquids in general decreases as altitude of a place increases where the external pressure is less than 1 atmosphere (normal b.p. of water is 373.15 K at 1 atmosphere) Vapour pressure 1atm Water Ethanol Temperature 76°C 100°C On top of Mount Everest, for example, where the atmospheric pressure is only about 260 mm Hg, water boils at approximately 71°C. Conversely, if the external pressure on a liquid is greater than 1 atm., the vapour pressure necessary for boiling than normal boiling is reached later, and the liquid boils at a temperature greater than normal boiling point. 1. Clausius-Clapeyron equation can be written in the following form : (A) P = Ae –ΔHvap /RT (C) 2. (B) ΔH vap dlog e P =– dT RT 2 ΔH vap dlog10 P = dT 2.303RT 2 (D) P = AeΔH vap /RT For a given liquid at a given temperature vapour pressure is given by : 400(K) + 10 T Vapour pressure of the liquid at 400 K is : log10 P (mm) = – (A) 9 mm 3. (C) 109 mm (B) –9 mm (D) 10–9 mm Latent heat of vaporisation of the above case in the given temperature range is : (A) –400 R (B) 400 R (C) –400 × 2.303 R (D) 400 × 2.303 R Comprehension # 4 Pressure (atm) U W V XY Z Temperature (°C) The phase diagram for a pure substance is shown above. Use this diagram and your knowledge about changes of phase to answer the following questions. 1. What does point V represent : (A) point of equilibrium (B) point of fusion (C) point of vaporisation (D) Triple point SOLUTION AND COLLIGATIVE PROPERTIES 2. What characteristics are specific to the system only at point V? (A) Liquid ‡ˆ ˆˆ ˆ† ˆˆ Solid (B) Solid ‡ˆ ˆˆ ˆ† ˆˆ Vapour (C) Liquid ‡ˆ ˆˆ ˆ† ˆˆ Vapour (D) Solid ‡ˆ ˆˆ ˆ† ˆˆ Liquid ‡ˆ ˆˆ ˆ† ˆˆ 3. Vapour What happens if temperature is increased from X to Y at 1.0 atm ? (A) solid is competely vaporised (B) solid and vapour are in equilibrium (C) solid and liquid are in equilibrium (D) liquid and vapour are in equilibrium 4. Select correct statement (s) : (A) curve VU is solid-liquid equilibrium curve (B) curve VU has a positive slope (C) curve VW is vapour pressure curve for liquid substance (D) In the solid - liquid mixture of the substance, solid will float 5. If the given substance is water then : (A) curve VU would have negative slope (B) in ice ‡ˆ ˆˆ ˆ† ˆˆ water liquid mixture, ice will float (C) as the temperature increases, pressure at which solid and liquid are in equilibrium, decreases (D) increase in pressure at constant temperature causes ice to be converted to liquid water 6. If the triple point pressure of a substance is greater than 1 atm, we expect : (A) the solid to sublime without melting (B) the boiling point temperature to be lower than the triple point temperature (C) the melting point of the solid to come at a lower temperature than the triple point (D) that the substance cannot exist as a liquid 7. In a phase change (say solid to liquid or liquid to solid) G = H – TS where : (A) H is the enthalpy change associated with making or breaking the intermolecular attractions that hold solid and liquid together and S is associated with change in disorder between the various phases. (B) H is associated with change in disorder while S is associated with energy change (C) both are associated with change in disorder (D) both are associated with change in energy EXERCISE - 3 : PART # I 1. 3. A q, s, B q, s, C q, s, D q, s A p; B p, q; C s; D p, r 2. A p, s, r, B p, q, t, C p, D p, q, t PART # II Comprehension # 1 : Comprehension # 2 : Comprehension # 3 : Comprehension # 4 : 1. B 1. B 1. A, B 1. D 2. 2. 2. 2. A D C D 3. 3. D B 4. A, B, C 5. A, B, C, D 6. A 7. A