19.5 g of fluoro acetic acid (molar mass = 78 g mol$$^{-1}$$) is dissolved in 500 g of water at 298 K. The depression in the freezing point was 1°C. What is K$$_a$$ of fluoro acetic acid? (For water, K$$_f$$ = 1.86 K kg mol$$^{-1}$$). Assume molarity and molality to have same values.

Moles of fluoro acetic acid dissolved:
$$n = \frac{19.5\ \text{g}}{78\ \text{g mol}^{-1}} = 0.25\ \text{mol}$$

Mass of water = 500 g = 0.5 kg.
Molality of the solution:
$$m = \frac{n}{\text{kg of solvent}} = \frac{0.25}{0.5} = 0.5\ \text{mol kg}^{-1}$$

The cryoscopic (freezing-point) equation is
$$\Delta T_f = i\,K_f\,m$$
Given $$\Delta T_f = 1\ ^\circ\text{C}$$ and $$K_f = 1.86\ \text{K kg mol}^{-1}$$,

$$1 = i \times 1.86 \times 0.5$$
$$\Rightarrow\ i = \frac{1}{1.86 \times 0.5} = \frac{1}{0.93} \approx 1.075$$

For a monoprotic weak acid $$\mathrm{HA \rightleftharpoons H^+ + A^-}$$,
the van’t Hoff factor is $$i = 1 + \alpha$$, where $$\alpha$$ is the degree of dissociation.
Therefore
$$\alpha = i - 1 = 1.075 - 1 = 0.075$$

Assuming density ≈ 1 g mL$$^{-1}$$, molarity ≈ molality, so $$C \approx 0.5\ \text{M}$$.

For a weak acid,
$$K_a = \frac{\alpha^2 C}{1 - \alpha}$$

Substituting the values:
$$\alpha^2 = (0.075)^2 = 0.0056$$
$$\alpha^2 C = 0.0056 \times 0.5 = 0.0028$$
$$1 - \alpha = 1 - 0.075 = 0.925$$
$$K_a = \frac{0.0028}{0.925} \approx 0.0030$$

$$K_a \approx 3 \times 10^{-3}$$

Option D which is: $$3 \times 10^{-3}$$