Solution and Colligative Properties Solved Examples

SOLUTION AND COLLIGATIVE PROPERTIES SOLVED EXAMPLES Ex.1 Sol. Ex.2 Sol. Ex.3 Sol. Mixture of volatile components A and B has total vapour pressure (in torr) : P = 254 – 119 xA Where xA is mol fraction of A in mixture. Hence PA° and PB° are (in torr) (A) 254, 119 (B) 119, 254 (C) 135, 254 (C) when xA = 0, xB = 1  P = PB°  PB° = 254, when xA = 1, xB = 0  PA° = P = 254 – 119 = 135 (D) 154, 119 When mango is placed in dilute aqueous solution of hydrochloric acid, it ? (A) Shrinks (B) Swells (C) Brusts (D) Nothing happens (A) The H+ ion concentration in the medium is lower than that of H+ concentration present in the mango juice, therefore osmotic pressure inside the mango is higher than that of aqueous hydrochloric acid. Therefore, H 2O will be passes out from the mango into aqueous hydrochloric acid. Therefore mango shrinks. Heptane and octane form ideal solution. At 373 K, the vapour pressures of the two liquids are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure, in bar, of a mixture of 25g of heptane and 35 g of octane ? (A) Heptane C7H16 mA = 100 (B) Octane C8H18 mB = 114 wA 25 = = 0.25 ; m A 100 0.25 = 0.45 xA = 0.25 + 0.30 = 0.45 nA = 35 = 0.3 114 0.3 xB = = 0.55 0.25 + 0.30 nB = p = p 0A x A  p 0B x B = 105.2 × 0.45 + 46.8 × 0.55 = 47.34 + 25.74 = 73.08 kPa Ex.4 Sol. A certain substance 'A' tetramerises in water to the extent of 80%. A solution of 2.5 g of A in 100 g of water lowers the freezing point by 0.3°C. The molar mass of A is : (A) 122 (B) 31 (C) 244 (D) 62 (D) 1– i 1 1– n 1– i ;i = 0.4 0.8 = 1 1– n T = iKf × m = 0.3 = 0.4 × 1.86 × w B ×1000 mB × w A 0.3 = 0.4 × 1.86 × 2.5 ×1000 m B ×100 mB = 62 Ex.5 Which of the following solutions will exhibit highest boiling point ? (A) 0.01 M Na2SO4 Sol. (B) 0.01 M KNO3 (C) 0.015 M urea (D) 0.015 M glucose (A) T = i × Kb × m i × m of Na2SO4 is highest, hence its boiling point will also be highest. Na2SO4 i × m = 3 × 0.01 = 0.03 KNO3 i × m = 2 × 0.01 = 0.02 Urea i × m = 1 × 0.015 = 0.015 Glucose i × m = 1 × 0.015 = 0.015 Ex.6 What is the osmotic pressure of 12% solution of can sugar (mol. wt. 342) at 17°C. Sol. 12 g. sugar is dissolved in 100 mL thus 342 g. sugar is dissolved in 8 litres Now, V = ST { n = 1} π= Ex.7 ST 0.0821× 290 = = 8.35 atm V 2.85 A dilute solution contains m mol of solute A in 1 kg of a solvent with molal elevation constant Kb. The solute dimerises in solution as 2A ‡ˆ ˆˆ ˆ† ˆˆ A2. Show that equilibrium constant for the dimer formation is K b (K b m – ΔTb ) (2ΔTb – K b m) 2 where Tb is the elevation of the boiling point for the given solution. Assume molarity = molality 2A ‡ˆ ˆˆ ˆ† ˆˆ A K= Sol. 2 m 0 mα 2 mα 2 (m – m) m (1 – ) initially after dimerization Where is the degree of dimerization and m is molality which is also molarity (given). Hence due to dimerization, final molality = m m(1 – ) + mα α  = m 1 –  2  2 i = 1 + (n – 1) 1  = 1 +  – 1 α 2   α  = 1 –   2 α  Tb = Kb × m × i = Kb × m 1 –   2 2(K b m – ΔTb ) Kbm equilibrium constant K for the dimer formation is   = SOLUTION AND COLLIGATIVE PROPERTIES mα α [A 2 ] K= = 2 2 2 = 2 2m(1 – α) 2 [A] m (1 – α)  K b m – ΔTb  K=2    Kbm   2(K b m – ΔTb )  2m 1 –  Kbm   K= K b (K b m – ΔTb ) (K b m – 2K b m + 2ΔTb )2 K= K b (K b m – ΔTb ) (2ΔTb – K b m) 2 2 Proved Ex.8 Distribution coefficient of an organic acid between water and benzene is 4.1 in favour of C6H6. If 5 g of acid is distributed in between 50 mL of benzene and 100 mL of water, calculate the concentration of acid in two solvents. Sol. Let the amount of organic acid in C6H6 layer = a g volume of C6H6 = 50 mL  concentration of acid in C6H6 = a g mL–1 50 since total amount of acid = 5 g  Amount of acid in H2O layer = (5 – a) g and volume of H2O = 100 mL 5–a g  Conc. of acid in H2O =    100  mL Now, K =  4.1 Conc.of acid in C6 H 6 Conc.of acid in H 2O = a 100 × 50 (5 – a) = a 100 × 50 (5 – a) or a = 3.361 g.  Amount of acid in 50 mL C6H6 = 3.361 g  acid concentration in C6H6 = Also, amount of acid in 100 mL  acid concentration in H2O = 3.361 ×1000 = 67.22 g/L 50 H2O = 5 – a = 5 – 3.361 = 1.639 g 1.639 ×1000 = 16.39 g/L 100 Ex.9 Sol. The freezing point depression of 0.001 m Kx[Fe(CN)6] is 7.10 × 10–3 K. Determine the value of x. Given, Kf = 1.86 K kg mol–1 for water. x = i × Kf × m 7.10 × 10–3 = i × 1.86 × 0.001 i = 3.817 = i –1 n –1 3.817 – 1 (x + 1) – 1 x = 2.817  3  Molecular formula of the compound is K3[Fe(CN)6]. 1= Ex.10 The composition of vapour over a binary ideal solution is determined by the composition of the liquid. If xA and yA are the mole fractions of A in the liquid and vapour, respectively find the value of xA for which (yA – xA) has maximum. What is the value of the pressure at this composition ? Sol. Since x A PA° PB° + (PA° – PB )x A Substracting xA from both the sides, we get yA = x A PA° – xA P + (PA° – PB° )x A Differentiating this with respect to xA, we get yA – xA = ° B x A PA°  PA° – PB°  PA° d(y A – x A ) = ° – 2 –1 PA + (PA° – PB° )x A PB° + (PA° – PB° )x A  dx A The value of xA at which yA – xA has a maximum value can be obtained by setting the above differential equal to zero. Thus, we have x A PA°  PA° – PB°  PA° – 2 –1 = 0 PA° + (PA° – PB° )x A PB° + (PA° – PB° )x A  PA° PB° – PB° PA° – PB The value of P at this composition is Solving for xA, we get x A = P = xA PA° + xB PB° or P = PB° + (PA° – PB°) xA  PA° PB° – PB°   or P = PB° + (PA° – PB°)  ° °  PA – PB  or P = PA° PB°