The conductivity of 0 · 2 M solution of KCl is 2 · 48 × 10^-2 S cm^-1. Calculate its molar conductivity and degree of dissociation (α). 3 Given:…

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