KBr is 80% ionized in solution. The freezing point of 0.4 molal solution of KBr is :
${\mathrm{K}}_{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}\right)=1.86\frac{\mathrm{Kkg}}{\mathrm{mole}}$

${\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{iK}}_{\mathrm{f}}\mathrm{m}$
$\underset{\underset{(1-\mathrm{\alpha })}{1}}{\mathrm{KBr}}⟶\underset{\underset{\mathrm{\alpha }}{0}}{{\mathrm{K}}^{+}}+\underset{\underset{\mathrm{\alpha }}{0}}{{\mathrm{Br}}^{-}}$
$\mathrm{i} = 1+\mathrm{\alpha }$
Here $\mathrm{\alpha } = 0.8 (80\%\mathrm{ionization}$)
i = 1.8
$\begin{array}{l}{\mathrm{\Delta T}}_{\mathrm{f}}=(1.8)(1.86)(0.4)=1.339\mathrm{K}\\ {\mathrm{T}}_{\mathrm{f}}=-1{.339}^{\circ }\mathrm{C}\end{array}$