Question

Calculate the freezing point of an aqueous solution of a non-electrolyte having osmotic pressure of 2.0 atm at 300 K.

(K_f = 1.86 K kg mol⁻¹, R = 0.0821 L atm K⁻¹ mol⁻¹)

Calculate the freezing point of an aqueous solution of a non-electrolyte having an osmotic pressure of 2.0 atm at 300 K.

Answer 1

Given: Osmotic pressure π = 2.0 atm

Temperature T = 300 K

R = 0.0821 L atm K⁻¹ mol⁻¹

K_f = 1.86 K kg mol⁻¹

Solvent = water, so freezing point of pure water = 273 K

πV = n RT

`n = (pi V)/(R * T)`

= `(2 xx 1)/(0.0821 xx 300)`

= `2/24.63`

n = 0.0812 mol/L

If the density of water is taken to be equal to 1, 1 litre of water will weigh 1000 g, i.e., 1 kg.

Hence, molality of the solution (m) = 0.0812

ΔT_f = K_f . m

= 1.86 × 0.0812

ΔT_f = 0.151 K

T_f = 273 − 0.151

T_f = 272.85 K 

∴ The freezing point of the solution is 272.85 K.