0.5 molal aqueous solution of a weak acid $$(HX)$$  is $$20\% $$  ionised. If $${k_f}$$  for water is $$1.86\,K\,kg\,mo{l^{ - 1}},$$    the lowering in freezing point of the solution is

Solution :
$$\eqalign{ & HX\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \rightleftharpoons \,\,\,{H^ + } + {X^ - } \cr & {\text{Weak acid}} \cr & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\left( {{\text{initially}}} \right) \cr & 1 - \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \,\,\,\,\left( {{\text{at equilibrium}}} \right) \cr & {\text{Total}} = 1 - \alpha + \alpha + \alpha \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \alpha \cr & \,\,\,\,\,\,\,\,\,\,i = \frac{{1 + \alpha }}{1} \cr & \,\,\,\,\,\,\,\,\alpha = 20\% \,{\text{dissociation}} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,\alpha = 0.2 \cr & i = 1 + \alpha \cr & \,\,\, = 1 + 0.2 \cr & \,\,\, = 1.2 \cr & \Delta {T_f} = i \times {K_f} \times m \cr & \,\,\,\,\,\,\,\,\,\,\, = 1.2 \times 1.86\,K\,kg\,mo{l^{ - 1}} \times 0.5 \cr & \,\,\,\,\,\,\,\,\,\,\, = 1.12\,K \cr} $$