At 300 K, 0.15 M sucrose solution is isotonic with 0.1 M KCl solution. Degree of dissociation of KCl in the solution is

For isotonic solution

$$\begin{gathered} {\mathrm{i}}_1{\mathrm{M}}_1\left(\mathrm{sucrose}\right)={\mathrm{i}}_2{\mathrm{M}}_2\left(\mathrm{KCl}\right) \\ 1\left(0.15\mathrm{M}\right)={\mathrm{i}}_2\left(0.1\mathrm{M}\right) \\ {\mathrm{i}}_2\left(\mathrm{KCl}\right)=\frac{0.15}{0.1}=1.5 \\ \underset{1-\mathrm{\alpha }}{\mathrm{KCl}}\leftrightharpoons \underset{\mathrm{\alpha }}{{\mathrm{K}}^{+}}+\underset{\mathrm{\alpha }}{{\mathrm{Cl}}^{-}} \\ \mathrm{i} = 1-\mathrm{\alpha }+\mathrm{\alpha }+\mathrm{\alpha }=1+\mathrm{\alpha } \\ 1+\mathrm{\alpha }=1.5 \\ \mathrm{\alpha }=1.5-1 \\ \mathrm{\alpha }=0.5; \end{gathered}$$