$\mathrm{The} \mathrm{molar} \mathrm{conductances} \mathrm{of} \mathrm{HC}(\mathrm{NaC} \mathrm{and} {\mathrm{CH}}_3 \mathrm{COONa} \mathrm{are} 426,126 \mathrm{and} 91{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}{\mathrm{mol}}^{-1} \mathrm{respectively}. \mathrm{The} \mathrm{molar} \mathrm{conductance} \mathrm{for} {\mathrm{CH}}_3 \mathrm{COOH} \mathrm{is}$

Given,

The molar Conductances of  $\mathrm{HC}\ell ,\mathrm{NaC}\ell \text{ and }{\mathrm{CH}}_3\mathrm{COONa}$ are 426, 126 and $91{\Omega }^{-1}{\text{cm}}^{\text{2}}{\text{mol}}^{\text{-1}}$ respectively.

The molar conductance for ${\text{CH}}_{\text{3}}\text{COOH=?}$

Applying Kohlrausch law,

$\therefore {\wedge }^{\text{o}}{\text{CH}}_{\text{3}}\text{COOH}={\wedge }^{\text{o}}{\text{CH}}_{\text{3}}\text{COONa}+{\wedge }^{\text{o}}\text{HC}\mathcal{l}-{\wedge }^{\text{o}}\text{NaC}\mathcal{l}$

$=91+426-126$

$=391{\Omega }^{-1}{\text{cm}}^{\text{2}}{\text{mol}}^{\text{-1}}$