The limiting molar conductance of $\mathrm{NaCl}, \mathrm{HCl}$ and $\mathrm{CH}_{3} \mathrm{COONa}$

Question

The limiting molar conductance of $\mathrm{NaCl}, \mathrm{HCl}$ and $\mathrm{CH}_{3} \mathrm{COONa}$ at infinite dilution are 126, 426 and $91 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ respectively. The liming molar conductance of $\mathrm{CH}_{3} \mathrm{COOH}$ will be\

Professor PC Thomas Classes · Answer

$\lambda_{0} H A c=\lambda_{0}[\mathrm{HCl}+\mathrm{NaAc}-\mathrm{NaCl}]$\ $=426+91-126=391$