In order to prepare one litre of a 1 N (norm | vedclass

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(B) The equivalent weight $(E)$ of $KMnO_4$ in an acidic medium is calculated as $E = \frac{	ext{Molar Mass}}{n	ext{-factor}}$.For $KMnO_4$ in an ac

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$31.6$

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(B) The equivalent weight $(E)$ of $KMnO_4$ in an acidic medium is calculated as $E = \frac{	ext{Molar Mass}}{n	ext{-factor}}$.For $KMnO_4$ in an acidic medium,the $n	ext{-factor}$ is $5$ (as $MnO_4^-$ is reduced to $Mn^{2+}$).Molar mass of $KMnO_4 = 39 + 55 + (4 	imes 16) = 158 \ g/mol$.Therefore,$E = \frac{158}{5} = 31.6 \ g/eq$.For a $1 \ N$ solution in $1 \ L$,the required mass is $N 	imes E 	imes V(L) = 1 	imes 31.6 	imes 1 = 31.6 \ g$.

In order to prepare one litre of a $1 \ N$ (normal) solution of $KMnO_4$,how many grams of $KMnO_4$ are required if the solution is used in an acidic medium for oxidation?

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