The molar conductances of $\mathrm{NaCl}$, $\mathrm{HCl}$ and $\mathrm{CH}{}_3\mathrm{COONa}$ at infinite dilution are $126.45$, $426.16$ and $91.0{\text{S cm}}^2{\text{mol}}^{-1}$ respectively. The molar conductance of $\mathrm{CH}{}_3\mathrm{COOH}$ at infinite dilution is. Choose the right option for your answer.

Answer: (D) $390.71{\text{S cm}}^2{\text{mol}}^{-1}$. \textbf{Answer:} (D) ${\Lambda }_m^{\circ }(\mathrm{CH}{}_3\mathrm{COOH})=390.71{\text{S cm}}^2{\text{mol}}^{-1}$ \textbf{Solution:} By Kohlrausch's law of independent migration of ions, the limiting molar conductivity of the weak electrolyte $\mathrm{CH}{}_3\mathrm{COOH}$ is built up from those of strong electrolytes: ${\Lambda }_m^{\circ }(\mathrm{CH}{}_3\mathrm{COOH})={\Lambda }_m^{\circ }(\mathrm{CH}{}_3\mathrm{COONa})+{\Lambda }_m^{\circ }(\mathrm{HCl})-{\Lambda }_m^{\circ }(\mathrm{NaCl})$ ${\Lambda }_m^{\circ }(\mathrm{CH}{}_3\mathrm{COOH})=91.0+426.16-126.45=390.71{\text{S cm}}^2{\text{mol}}^{-1}$ (The $\mathrm{Na}{}^{+}$ and $\mathrm{Cl}{}^{-}$ contributions cancel, leaving ${\lambda }^{\circ }(\mathrm{CH}{}_3\mathrm{COO}{}^{-})+{\lambda }^{\circ }(\mathrm{H}{}^{+})$.)