2 molal solution of a weak acid HA has a freezing point of 3.885°C. The degree of dissociation of this acid is ________ $$\times 10^{-3}$$. (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]

The depression in freezing point is given by $$\Delta T_f = i \cdot K_f \cdot m$$, where $$i$$ is the van't Hoff factor, $$K_f$$ is the molal depression constant, and $$m$$ is the molality.

The freezing point of the solution is $$-3.885°\text{C}$$ (since pure water freezes at $$0°\text{C}$$), so $$\Delta T_f = 0 - (-3.885) = 3.885°\text{C}$$.

Substituting the known values: $$3.885 = i \times 1.85 \times 2$$, which gives $$i = \frac{3.885}{3.70} = 1.050$$.

For a weak acid HA that dissociates as $$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$$, the van't Hoff factor is related to the degree of dissociation $$\alpha$$ by $$i = 1 + \alpha$$, since one molecule dissociates into two particles.

Therefore $$1 + \alpha = 1.050$$, giving $$\alpha = 0.050 = 50 \times 10^{-3}$$.

The answer is $$50$$.