Question

An aqueous solution of a non-volatile solute freezes at 272.4 K, while pure water freezes at 273.0 K. Determine the following:

(Given K_f = 1.86 K kg mol⁻¹, K_b = 0.512 K kg mol⁻¹ and vapour pressure of water at 298 K = 23.756 mm of Hg)

1. The molality of solution
2. Boiling point of solution
3. The lowering of vapour pressure of water at 298 K.

Answer 1

Given: Freezing point of solution = 272.4 K

Freezing point of pure water = 273.0 K

ΔT_f = 273.0 − 272.4 = 0.6 K

K_f = 1.86 K kg mol⁻¹

K_b = 0.512 K kg mol⁻¹

Vapour pressure of pure water at 298 K = 23.756 mm Hg

1. ΔT_f = K_f m

`m = (Delta T_f)/K_f`

= `0.6/1.86`

= 0.3226 mol/kg

2. ΔT_b = K_b m

= 0.512 × 0.3226

= 0.1652 K

Boiling point = 373.0 + 0.1652

= 373.17 K

3. Relative lowering of vapour pressure is given by

`(Delta P)/P^circ = chi_"solute"`

= `n_"solute"/n_"solvent"`

Assuming 1 kg (1000 g) of water:

Moles of water = `1000/18 = 55.56`

Moles of solute = molality × kg of solvent = 0.3226 mol

`chi_"solute" = 0.3226/(0.3226 + 55.56)`

= 0.00578

Now

`Delta P = P^circ * chi_"solute"`

= 23.756 × 0.00578

= 0.1373 mm Hg