What is degree of dissociation of $\mathrm{CH}_3 \mathrm{COOH}$ if $\wedge^{\circ}\left(\mathrm{CH}_3…

The degree of dissociation $\alpha$ for acetic acid is determined from the ratio of its molar conductivity at concentration $c = 5 \times 10^{-2}\,\mathrm{M}$ to its limiting molar conductivity: $\alpha = \frac{\wedge_m}{\wedge_m^{\circ}}$.

Using Kohlrausch's law, the limiting molar conductivity is the sum of the limiting ionic conductivities: $\wedge_m^{\circ}(\mathrm{CH}_3\mathrm{COOH}) = \wedge^{\circ}(\mathrm{CH}_3\mathrm{COO}^{-}) + \wedge^{\circ}(\mathrm{H}^{+}) = 50 + 350 = 400\,\mathrm{S\,cm}^2\,\mathrm{mol}^{-1}$.

Given $\wedge_m = 20\,\mathrm{S\,cm}^2\,\mathrm{mol}^{-1}$, we find $\alpha = \frac{20}{400} = 0.05 = 5 \times 10^{-2}$.

The value corresponds to option C.