Calculate the equivalent weight of $KMnO_4$ in acidic,basic,and neutral media.

(N/A) The equivalent weight is calculated as $E = \frac{M}{n}$,where $M$ is the molar mass and $n$ is the change in oxidation state (number of electrons gained).
$1$. Acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. Here,$n = 5$,so equivalent weight is $\frac{M}{5}$.
$2$. Basic medium: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$. Here,$n = 3$,so equivalent weight is $\frac{M}{3}$.
$3$. Neutral or weakly basic medium: $MnO_4^- + e^- \rightarrow MnO_4^{2-}$. Here,$n = 1$,so equivalent weight is $\frac{M}{1}$.