Chemistry - Liquid Solution Question with Solution | TestHub

In acidic medium, $\begin{array}{rl}& {\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5e^{-}⟶{\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}\\ & \text{ (From }{\mathrm{KMnO}}_4\text{ ) }\end{array}$ Hence, equivalent weight of ${\mathrm{KMnO}}_4$ in acidic medium $=\frac{\text{ Moleculer weight }}{5}=\frac{158}{5}=31.6$ $\therefore 1\mathrm{N}$ solution can be prepared by dissolving 31.6 g ${\mathrm{KMnO}}_4$ in 1L solution.