# Equivalent conductivity of $Fe_2(SO_4)_3$ is related to molar conductivity by the expression

$\begin{array}{1 1}\Lambda_{eq}=\Lambda_m\\\Lambda_{eq}=\Lambda_m/3\\\Lambda_{eq}=3\Lambda_m\\\Lambda_{eq}=\Lambda_m/6\end{array}$

Answer : $\Lambda_{eq}=\Lambda_m/6$
Since total charge of cations is 6+ in $Fe_2(SO_4)_3$,we have
$\Lambda_{eq}(Fe_2(SO_4)_3)=(1/6)\Lambda_m(Fe_2(SO_4)_3)$