The freezing point of 0.2 molal ${\mathrm{K}}_2{\mathrm{SO}}_4$ is $-{1.1}^{o}C$. Calculate percentage degree of dissociation of ${\mathrm{K}}_2{\mathrm{SO}}_4$. ${\mathrm{K}}_{\mathrm{f}}$ for water is $1.{86}^o$

${\mathrm{\Delta T}}_{\mathrm{f}}=$freezing point of water - freezing point of solution $=0^{\mathrm{o}}\mathrm{C}-(-1{.1}^{\mathrm{o}}\mathrm{C})=1{.1}^{\mathrm{o}}$

We know that, 

$\begin{array}{l}{\mathrm{\Delta T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}\\ 1.1=\mathrm{i}\times 1.86\times 0.2\\ \therefore \mathrm{i}=\frac{1.1}{1.86\times 0.2}=2.95\end{array}$

But we know $\mathrm{i}=1+(\mathrm{n}-1)\mathrm{\alpha }$

$\begin{array}{l}2.95=1+(3-1)\mathrm{\alpha }=1+2\mathrm{\alpha }\\ \mathrm{\alpha }=0.975\end{array}$

Percentage degree of dissociation = 97.5