Chemistry - Electrochemistry Question with Solution | TestHub

By Kohlraush's law, ${\wedge }_{\text{eq }}^0\mathrm{NaCl}=126.45$ $\begin{array}{rl}& {\lambda }_{{\mathrm{Na}}^{+}}^0+{\lambda }_{{\mathrm{Cl}}^{-}}^0=126.45\\ & {\lambda }_{{\mathrm{H}}^{+}}^0+{\lambda }_{{\mathrm{Cl}}^{-}}^0=\mathrm{426.16.}\\ & {\lambda }_{{\mathrm{CH}}_3{\mathrm{OO}}^{-}}^0+{\lambda }_{{\mathrm{Na}}^{+}}^0=91\end{array}$ on adding (2) and (3) then subtract (1) from it $\begin{array}{rl}& {\lambda }_{{\mathrm{CH}}_3{\mathrm{COO}}^{-}}^0+{\lambda }_{{\mathrm{Na}}^{+}}^0=517.16-126.45\\ & {\wedge }_{\left({\mathrm{CH}}_3\mathrm{COOH}\right)}^0=390.71{\mathrm{ohm}}^{-1}{\mathrm{cm}}^2\end{array}$