The freezing point of a solution of $4.0 \mathrm{g} {\mathrm{CdI}}_2$ in$95 \mathrm{g}$ of water was ${0.25}^{\circ }\mathrm{C}$. Calculate the degree of dissociation of salt (nearest integer value)$\text{ ( }{K}_f\text{ for water }=1.{86}^{\circ }\mathrm{C}/\mathrm{m};\text{ (At. wt. }\mathrm{Cd}=112\text{; }$ $\mathrm{I}=127)$

$$\begin{gathered} \mathrm{wt}. \mathrm{of} {\mathrm{CdI}}_2\left(\mathrm{w}\right)=4.0 \mathrm{g} \\ \mathrm{wt}. \mathrm{of} \mathrm{water}\left(\mathrm{W}\right)=95 \mathrm{g} \\ \mathrm{mol}.\mathrm{wt}. \mathrm{of} {\mathrm{CdI}}_2\left(\mathrm{m}\right)=112 + 127\times 2=366 \\ ∆{\mathrm{T}}_{{\mathrm{f}}_{\mathrm{cal}}}=\frac{{\mathrm{K}}_{\mathrm{f}}\times 1000\times \mathrm{w}}{\mathrm{W}\times \mathrm{m}}=\frac{1.86\times 1000\times 4.0}{95\times 366} \\ =0.2140°\mathrm{C} \\ ∆{\mathrm{T}}_{{\mathrm{f}}_{\mathrm{obs}}}=0°\mathrm{C} +0.25°\mathrm{C}=0.25°\mathrm{C} \\ \mathrm{van}\text{'}\mathrm{t} \mathrm{hoff} \mathrm{factor}\left(\mathrm{i}\right)=\frac{∆{\mathrm{T}}_{{\mathrm{f}}_{\mathrm{obs}}}}{∆{\mathrm{T}}_{{\mathrm{f}}_{\mathrm{cal}}}}=\frac{0.25}{0.2140} =1.168 \\ \mathrm{The} \mathrm{degree} \mathrm{of} \mathrm{dissociation}\left(\mathrm{\alpha }\right) \mathrm{of} {\mathrm{CdI}}_2=\frac{\mathrm{i}-1}{\mathrm{n}-1} \\ \mathrm{where} \mathrm{n}= \mathrm{no}. \mathrm{of} \mathrm{ions}=3 \mathrm{for} {\mathrm{Cd}}^{2+} \mathrm{and} 2{\mathrm{I}}^{-} \mathrm{ions} \\ \therefore \mathrm{\alpha }=\frac{1.168-1}{3-1}=0.084 \\ \therefore \mathrm{\alpha }\cong 8\% \end{gathered}$$