$$HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$$. The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20°C. The dissociation constant for the acid is : (Given: $$K_f(H_2O) = 1.8$$ K kg mol$$^{-1}$$, molality $$\equiv$$ molarity)

The acid dissociates according to $$HA \; \rightleftharpoons \; H^{+} + A^{-}$$

For a colligative-property experiment we use the van’t Hoff factor $$i$$.
The freezing-point depression is related to $$i$$ by the formula
$$\Delta T_f = i \, K_f \, m$$ $$-(1)$$

Data given: $$\Delta T_f = 0.20^{\circ}\text{C}$$, $$K_f = 1.8 \text{ K kg mol}^{-1}$$, molality $$m = 0.1 \text{ m}$$.
Substituting in $$(1)$$:

$$0.20 = i \,(1.8)\,(0.1)$$

$$\Rightarrow \; i = \frac{0.20}{1.8 \times 0.1} = \frac{0.20}{0.18} = 1.11$$ (keep three significant figures)

For a weak monobasic acid each formula unit can produce two ions, so $$n = 2$$.
The relation between $$i$$ and the degree of dissociation $$\alpha$$ is
$$i = 1 + \alpha (n-1) = 1 + \alpha$$ $$-(2)$$

Using $$i = 1.11$$ in $$(2)$$:
$$\alpha = i - 1 = 1.11 - 1 = 0.11$$

The equilibrium concentration of the acid is $$c = 0.1 \text{ mol L}^{-1}$$.
For a weak monoprotic acid the dissociation constant is
$$K_a = \frac{c \alpha^{2}}{1 - \alpha}$$ $$-(3)$$

Substituting the values:
$$K_a = \frac{0.1 \times (0.11)^{2}}{1 - 0.11} = \frac{0.1 \times 0.0121}{0.89} = \frac{0.00121}{0.89} \approx 1.4 \times 10^{-3}$$

Therefore $$K_a \approx 1.38 \times 10^{-3}$$.

Option A is correct.