Question

An aqueous solution freezes at 272.4 K, while pure water freezes at 273.0 K. Determine

1. the molality of the solution
2. boiling point of the solution
3. lowering of vapour pressure of water at 298 K.

(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 Hg)

Answer 1

Given: Freezing point of solution (T_f) = 272.4 K

Freezing point of pure water `(T_f^circ)` = 273.0 K

Depression in freezing point:

`Delta T_f = T_f^circ − T_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 water at 298 K (P°) = 23.756 mm Hg

i. Molality of the solution:

ΔT_f = K_f . m

⇒ `m = (Delta T_f)/(K_f)`

= `0.6/1.86`

m = 0.3226 mol/kg

ii. Boiling point of the solution:

ΔT_b = K_b . m

= 0.512 × 0.3226

= 0.165 K

T_b = 373 + ΔT_b

= 373 + 0.165

= 373.165 K

iii. Lowering of vapour pressure:

Relative lowering of vapour pressure `((P^circ - P)/P^circ)` = mole fraction of solute

= `m/(m + 55.5)`

= `0.3226/(0.3223 + 55.5)`

= 0.00578

`Delta P = P^circ * (P^circ - P)/P^circ`

= 23.756 × 0.00578

= 0.1373 mm Hg