The conductivity of $\displaystyle 0 \cdot 2 \mathrm{M}$ solution of KCl is $\displaystyle 2 \cdot 48 \times 10^{-2} \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its molar conductivity and degree of dissociation $\displaystyle (\alpha)$. $\displaystyle 3$ Given : \[\begin{aligned} \lambda_{\mathrm{K}^{+}}^{\mathrm{o}} & =73 \cdot 5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \\ \lambda_{\mathrm{Cl}^{-}}^{\mathrm{o}} & =76 \cdot 5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \end{aligned} \]

\[\begin{aligned} \mathbf{\Lambda}_{m} & =\mathrm{k} / \mathrm{C} \\ \mathbf{\Lambda}_{m} & =\frac{\mathrm{k} \times 1000}{\mathrm{M}} \\ & =\frac{1000 \mathrm{~cm}^{3} / \mathrm{L} \times 2.48 \times 10^{-2} \mathrm{Scm}^{-1}}{0.2 \mathrm{~mol} \mathrm{~L}^{-1}} \\ & =124 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \\ \mathbf{\Lambda}_{m}^{\circ} & =\lambda_{+}^{\circ}+\lambda_{-}^{\circ} \\ & =(73.5+76.5) \mathrm{S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \\ & =150 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \\ \alpha & =\mathbf{\Lambda}_{m} / \mathbf{\Lambda}^{\circ} m \\ & =\frac{124 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}{150 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}} \\ & =0.826 \text { (approx.) } \end{aligned} \]