Question

2 molal solution of a weak acid HA has a freezing point of 3.885°C. The degree of dissociation of this acid is ______ × 10⁻³. (Round off to the nearest integer).

(Given: Molal depression constant of water = 1.85 K kg mol⁻¹, Freezing point of pure water = 0°C)

Answer 1

2 molal solution of a weak acid HA has a freezing point of 3.885°C. The degree of dissociation of this acid is 50 × 10⁻³.

Explanation:

Given: Molality (m) = 2 mol/kg

Freezing point of a solution = −3.885°C

Freezing point of pure water = 0°C

Depression in freezing point ΔT_f = 0 − (−3.885) = 3.885°C

K_f (cryoscopic constant) = 1.85 K kg mol⁻¹

Let the degree of dissociation of the weak acid be α.

For a weak monoprotic acid HA (dissociates into H⁺ and A⁻), total number of particles = 1 + α

So, Van’t Hoff factor (i) = 1 + α

Now, ΔT_f ​= i ⋅ K_f ​⋅ m

3.885 = (1 + α) × 1.85 × 2

3.885 = 3.7(1 + α)

`3.885/3.7 = 1 + alpha`

1.05 = 1 + α

α = 0.05

α = 50 × 10⁻³