The equivalent weight of potassium permangana | vedclass

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(B) The equivalent weight of an oxidizing agent is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molar mass}}{	ext{n-factor}}$.For $KMnO_4$

Vedclass · Accepted Answer

$31.6$

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(B) The equivalent weight of an oxidizing agent is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molar mass}}{	ext{n-factor}}$.For $KMnO_4$ in an acidic medium,the reduction reaction is: $MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$.The change in oxidation state of $Mn$ is from $+7$ to $+2$,so the $n$-factor is $5$.The molar mass of $KMnO_4$ is $158 \ g/mol$.Therefore,$	ext{Equivalent weight} = \frac{158}{5} = 31.6$.

The equivalent weight of potassium permanganate $(KMnO_4)$ in an acidic solution is:

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