(D) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $K_2MnO_4$.The reaction is: $MnO_4^- + e^- \to MnO_4^{2-}$.The change

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(D) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $K_2MnO_4$.The reaction is: $MnO_4^- + e^- \to MnO_4^{2-}$.The change in oxidation state of $Mn$ is from $+7$ to $+6$,so the $n$-factor is $1$.Equivalent weight = $\frac{\text{Molecular weight}}{n\text{-factor}} = \frac{158}{1} = 158$.