The number of moles of electrons required to reduce $20 \mathrm{~cm}^{3}$ of 0.05 M solution of $\ma

Question

The number of moles of electrons required to reduce $20 \mathrm{~cm}^{3}$ of 0.05 M solution of $\mathrm{KMnO}_{4}$ to manganese sulphate in acid medium is\

Professor PC Thomas Classes · Answer

Millimoles of $\mathrm{KMnO}_{4}=20 	imes 0.05 \mathrm{~m}$ moles $=1 \mathrm{~m}$ mole $=1 	imes 10^{-3}$ moles +7 \ 1 mole of $\mathrm{KMnO}_{4}$ requires 5 moles of $\mathrm{e}^{-}$\ $	herefore$ For $1 	imes 10^{-3}$ moles, $5 	imes 1 	imes 10^{-3}$ moles of $\mathrm{e}^{-}$\