Molecular weight of $KMn{O}_4$ in acidic medium and neutral medium will be respectively :

The correct option is B $5\times \text{equivalent weight and}3\times \text{equivalent weight}$
$\text{Equivalent weight}=\frac{\text{Molecular~weight}}{\text{No. of electrons involved in reaction}}$
$\text{Molecular weight= Equivalent~weight}\times \text{No. of electrons involved in reaction}$
$KMn{O}_4$ in acidic medium,
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$
Number of electrons involved = 5
Therefore, $\text{Molecular weight}=5\times \text{Equivalent weight}$
$KMn{O}_4$ in basic medium,
$Mn{O}_4^{-}+e^{-}\to Mn{O}_4^{2-}$
$Mn{O}_4^{2-}+2e^{-}\to Mn{O}_2$
Number of electrons involved = 3
Therefore, $\text{Molecular weight}=3\times \text{Equivalent weight}$