The equivalent weight of ${\mathrm{KMnO}}_4$ in acidic medium is

The balanced chemical equation for the oxidation of ${\mathrm{KMnO}}_4$ in acidic medium is given below. 

$2{\mathrm{KMnO}}_4+3{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+3{\mathrm{H}}_2\mathrm{O}+2{\mathrm{MnSO}}_4+5\left(\mathrm{O}\right)$

In the above equation, the net change is as given below.

${\mathrm{Mn}}^{+7}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{+2}$

Equivalent weight of  ${\mathrm{KMnO}}_4=\frac{\mathrm{molecular} \mathrm{weight} \mathrm{of} {\mathrm{KMnO}}_4}{\mathrm{No} \mathrm{of} \mathrm{electrons} \mathrm{transfered}}$

Note: Molecular weight of ${\mathrm{KMnO}}_4 \mathrm{is} 158\mathrm{g}/\mathrm{mol}$.