A 0.1 molal aqueous solution of a weak acid is 30 % ionized. If ${\mathrm{K}}_{\mathrm{f}}$ for water is 1.86 °C/m, the freezing point of the solution will be

Molality, $\mathrm{m} = 0.1$; vant Hoff's factor, i for a weak acid $=1+\alpha$ where $\mathrm{\alpha }=$ degree of dissociation $= 30 \%$

$=30/100=0.3$ 

But $\mathrm{i}=1+\mathrm{\alpha }.\text{ So },\mathrm{i}=1+0.3=1.3$ 

Thus, ${\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{iK}}_{\mathrm{f}}\times \mathrm{m}=1.3\times 1.86\times 0.1=0{.24}^{\circ }\mathrm{C}$

$\therefore$ Freezing point $=0-{0.24}^{\circ }\mathrm{C}=-{0.24}^{\circ }\mathrm{C}$