Explanation

(d)
In acidic medium following reaction takes place.
$8H^{+} + 5e^{-} + MnO_{4}^{-} \rightarrow Mn^{2 +} + 4H_{2}O$
$\therefore$ Equivalent weight of $KMnO_{4}$ in acidic medium
$= \frac{molecular\ weight\ of\ KMnO_{4}}{5}$
$= \frac{158}{5} = 31.6$