Freezing Point Depression:

$ \Delta T_f = i \times K_f \times m $

Given:

$ \Delta T_f = 0.2 \, \text{°C}, \quad K_f = 1.8 \, \text{K kg mol}^{-1}, \quad m = 0.1 \, \text{m} $

Substituting the given values:

$ 0.2 = i \times 1.8 \times 0.1 $

Solving for $i$:

$ i = \frac{0.2}{1.8 \times 0.1} = \frac{20}{18} = \frac{10}{9} $

Degree of Dissociation ($\alpha$):

For the reaction $\mathrm{HA} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-}$:

$ i = 1 + \alpha $

Given $i = \frac{10}{9}$:

$ \frac{10}{9} = 1 + \alpha $

$ \alpha = \frac{1}{9} $

Dissociation Constant ($K_{eq}$):

$ \mathrm{K}_{eq} = \frac{[H^+][A^-]}{[HA]} $

At equilibrium:

$ [H^+] = [A^-] = \alpha \times C = \frac{1}{9} \times 0.1 $

$ [HA] = 0.1 \times (1 - \alpha) = 0.1 \times \left(1 - \frac{1}{9}\right) $

Substituting these into $K_{eq}$:

$ \mathrm{K}_{eq} = \frac{(0.1 \times \frac{1}{9})^2}{0.1 \times \left(1 - \frac{1}{9}\right)} $

Simplifying:

$ \mathrm{K}_{eq} = \frac{0.1 \times \left(\frac{1}{81}\right)}{0.1 \times \frac{8}{9}} = \frac{1}{720} $

Therefore:

$ \mathrm{K}_{eq} = 1.38 \times 10^{-3} $