For the weak acid $\text{HX}$, the ionization reaction is: $\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^-$ The number of ions produced per molecule upon dissociation, $n = 2$. Given, the degree of ionization $\alpha = 20\% = 0.2$.

The van't Hoff factor ($i$) is calculated as: $i = 1 + \alpha(n - 1) = 1 + 0.2(2 - 1) = 1.2$

The depression (lowering) in freezing point is given by the formula: $\Delta T_f = i \times K_f \times m$

Given values: Molality, $m = 0.5 \text{ mol kg}^{-1}$ Molal depression constant, $K_f = 1.86 \text{ K kg mol}^{-1}$

Substituting the values: $\Delta T_f = 1.2 \times 1.86 \times 0.5 = 1.116 \text{ K} \approx 1.12 \text{ K}$.

Therefore, the lowering in freezing point of the solution is $1.12 \text{ K}$.