$\mathrm{KBr}$ is $80 \%$ ionized in solution. The freezing point of $0.4$ molal solution of $\mathrm{KBr}$ is :
$\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}ight)=1.86 \mathrm{~K} \mathrm{~kg} / \mathrm{mole}$

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Aicharya · Accepted Answer

$\Delta T_f=i K_f m$ $\begin{array}{rcc} & \mathrm{KBr} & ightarrow \mathrm{K}^{+} & +\mathrm{Br}^{-} \ & 1 & 0 & 0 \ & (1-\alpha) & \alpha & \alpha \end{array}$ $i=1+\alpha$ Here $\alpha=0.8$ ($80 \%$ ionization) $\begin{aligned} & i=1.8 \ & \Delta T_f=(1.8)(1.86)(0.4)=1.339 K \ & T_f=-1.339^{\circ} \mathrm{C} \end{aligned}$ *

$\mathrm{KBr}$ is $80 \%$ ionized in solution. The freezing point of $0.4$ molal solution of $\mathrm{KBr}$…

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