In $\mathrm{KMnO}_{4}$ titration in acidic medium, equivalent mass of $\mathrm{KMnO}_{4}$ is\

Question

Professor PC Thomas Classes · Answer

(b) : In $\mathrm{KMnO}_{4}$ titration, $\mathrm{KMnO}_{4}$ is reduced as :\ $\mathrm{MnO}_{4}{ }^{-}+8 \mathrm{H}^{+}+5 e^{-} ightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$\ Eq. mass $=\frac{	ext { Mol. mass }}{	ext { No. of electrons gain or lost }}=\frac{M}{5}$