Question

The relative lowering of vapour pressure of an aqueous solution of a non volatile solute of molecular weight 60 is 0.018. If kf of water is 1.86 degree per metre the depression in freezing point will be

Answer 1

Given:

Relative lowering of vapour pressure = 0.018

Molecular weight = 60 gm

Kf = 1.86

To Find:

Depression in freezing point.

Calculation:

- We know that

Relative lowering of vapour pressure = Mole Fraction of solute

⇒ (P₀-P) / P = (n + N) /n = 0.018

⇒ n / (n + N) = 1 / 0.018

⇒ N / n = (1 / 0.018) - 1

⇒ N / n = 0.982 / 0.018

- Now, molality = (n / N) × 1000/18

⇒ m = (0.018 / 0.982) × 1000/18

⇒ m = 1.0183

- Now, depression in freezing point = Kf × m

⇒ ΔTf = 1.86 × 1.0183

⇒ ΔTf = 1.894 °C

- So the depression in freezing point is 1.894 °C.