The ionization constnat of a weak acid is 1.6 × 10^–5 and the molar conductivity at infinite dilution is 380 × 10^–4 sm²mol^–1. If the cell constant is 0.01m^–1, then conductance of 0.01M acid solution is

$\begin{array}{l}{\mathrm{K}}_{\mathrm{a}}={\mathrm{C\alpha }}^2;\\ \mathrm{\alpha }=\sqrt{{\mathrm{K}}_{\mathrm{a}}/\mathrm{C}}=\sqrt{1.6\times {10}^{-5}/{10}^{-2}}\\ =\sqrt{1.6\times {10}^{-3}}=\sqrt{16\times {10}^{-4}}\\ \mathrm{\alpha }=4\times {10}^{-2}=0.04;\mathrm{\alpha }={\mathrm{\Lambda }}_{\mathrm{M}}/{\mathrm{\Lambda }}_{\mathrm{M}}^0\\ \Rightarrow {\wedge }_{\mathrm{M}}=\mathrm{\alpha }\times {\wedge }_{\mathrm{M}}^0=0.04\times 380\times {10}^{-4}\\ \Rightarrow {\wedge }_{\mathrm{M}}=\mathrm{K}\times 1000/\mathrm{M}\end{array}$
$\begin{array}{l}\mathrm{K}=\frac{{\wedge }_{\mathrm{M}}\times \mathrm{M}}{1000}=\frac{0.04\times 380\times {10}^{-4}\times 0.01}{1000}\\ \mathrm{K}=0.152\times {10}^{-7}=1.52\times {10}^{-8}\\ \mathrm{C}=\mathrm{K}/\mathrm{G}=1.52\times {10}^{-8}/0.01=1.52\times {10}^{-6}\\ \mathrm{C}=1.52\times {10}^{-6}\mathrm{S}-{\mathrm{m}}^3=1.52{\mathrm{SCm}}^3\end{array}$