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# If one molecule of an electrolyte dissociates to produce $v_+$ number of cations $A^{z+}$ and $v_-$ number of anions $B^{z-}$,then the relation between molar conductivity $(\Lambda_m)$ and equivalent conductivity $(\Lambda_{eq})$ is

$\begin{array}{1 1}\Lambda_m=(v_+z_+)\Lambda_{eq}\\\Lambda_m=\Lambda_{eq}(v_+z_+)\\\Lambda_m=v_+\Lambda_{eq}/z_+\\\Lambda_m=z_+\Lambda_{eq}/v_+\end{array}$

Answer : $\Lambda_m=(v_+z_+)\Lambda_{eq}$
The relation connecting molar and equivalent conductivities is $\Lambda_m=(v_+z_+)\Lambda_{eq}$