If M is the molecular weight of KMnO_4,what w | vedclass

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(A) The chemical reaction for the conversion of $KMnO_4$ to $K_2MnO_4$ is:$KMnO_4 + e^- \rightarrow K_2MnO_4$ (in basic medium).In $KMnO_4$,the oxidat

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$M$

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(A) The chemical reaction for the conversion of $KMnO_4$ to $K_2MnO_4$ is:$KMnO_4 + e^- ightarrow K_2MnO_4$ (in basic medium).In $KMnO_4$,the oxidation state of $Mn$ is $+7$.In $K_2MnO_4$,the oxidation state of $Mn$ is $+6$.The change in oxidation state is $|7 - 6| = 1$.Thus,the number of electrons gained per molecule is $1$.Equivalent weight = $\frac{	ext{Molecular weight}}{	ext{n-factor}} = \frac{M}{1} = M$.

If $M$ is the molecular weight of $KMnO_4$,what will be its equivalent weight when it is converted into $K_2MnO_4$?

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