(a) Draw the graph between vapour pressure and temperature andexplain the elevation in boiling point of a solvent in solution.

(b) Determine the osmotic pressure of a solution prepared by

dissolving 25 mg of K₂SO₄ in 2 litres of water at 25°C assuming it to be completely dissociated. (Atomic masses K = 39 u, S = 32 u, O = 16 u)

OR

(a) Write two characteristics of non-ideal solution.

(b) 2 g of benzoic (C₆H₅COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol^–1. What is the percentage association of acid if it forms dimer in solution ?

(a)

When a solute is added to a solvent, the vapour pressure of the solvent decreases and at higher temperatures, it becomes equal to atmospheric pressure.

Boiling point elevation occurs whenever a non-volatile solute is added to a pure solvent.

This phenomenon occurs when the boiling point of the solvent is increased when another compound, a solute is added, such that the solution, so formed has a higher boiling point than the pure solvent.

rT_b = K_b × m

Where rT_b the elevation in boiling point,

K_b is the molal boiling point elevation constant,

m is the molality of solute.

(b) K₂SO₄ dissociates as:

Total ions produced (n) = 2+1 = 3

C = molarity of solute =

Molecular mass = 2×39 + 32 + 16×4 = 174 gm

Mass given = 25mg = 0.025gm

Moles =

Volume in L = 2L

∴ C =[M]

R, ideal gas constant = 0.0821L atm mol^-1K^-1.

Temperature T = 25° C = (25+273)K = 298K

Osmotic pressure π = n × C × R × T

= (3 × × 0.0821 × 298)atm

=5.27×10^-3 atm.

Therefore, osmotic pressure is 5.27×10^-3 atm.

OR

(a) Some characteristics of non-ideal solution:

(i)Volume change of mixing should not be zero.

(ii)Heat change on mixing should not be zero.

(iii)Solute molecules dissociate in the ideal solution.

(iv)Solute molecules associate in the ideal solution.

(v)Ideal solutions must not obey Raoult’s law at all concentrations.

Eg.Mixture of acetone and chloroform.

(b) rT_f = i× K_f × m

Where rT_fthe depression in freezing point = 1.62 K

i is the Vant-Hoff factor

K_f is the molal freezing point depression constant = 4.9 K kg mol^–1.

m is the molality of solute.

As the acid dimerises, N = 2

Molar mass of CH₃COOH = 61gm, when it dimerises, it becomes (2×61)gm = 122gm.

Moles of acid (solute) = 2/122

Weight of solvent in kg = 0.025kg.

So, m = =0.656

Thus, i = =0.504

Degree of association α = N(1-i)

Replacing the value of N=2 and i = 0.504, we get:

α = 2(1-0.504) = 0.992 = (0.992×100)% = 99.2%

So, 99.2% association.