In alkaline condition,KMnO_4 reacts as 2KMnO_ | vedclass

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(A) The molecular mass of $KMnO_4$ is $39 + 55 + (4 	imes 16) = 158 \ g/mol$.In the given reaction,the oxidation state of $Mn$ changes from $+7$ in $

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$158$

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(A) The molecular mass of $KMnO_4$ is $39 + 55 + (4 	imes 16) = 158 \ g/mol$.In the given reaction,the oxidation state of $Mn$ changes from $+7$ in $KMnO_4$ to $+6$ in $K_2MnO_4$.The change in oxidation state is $7 - 6 = 1$.Therefore,the $n$-factor for $KMnO_4$ is $1$.The equivalent weight is calculated as $E = \frac{	ext{Molecular Mass}}{n	ext{-factor}} = \frac{158}{1} = 158$.

In alkaline condition,$KMnO_4$ reacts as $2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + [O]$. The equivalent weight of $KMnO_4$ would be (Atomic mass of $K = 39$,$Mn = 55$,$O = 16$).

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