M is the molecular weight of KMnO_4. The equi | vedclass

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(A) The chemical reaction for the conversion is: $\mathop {KMnO_4}\limits^{+7} 	o \mathop {K_2MnO_4}\limits^{+6}$.The change in the oxidation state o

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$M$

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(A) The chemical reaction for the conversion is: $\mathop {KMnO_4}\limits^{+7} 	o \mathop {K_2MnO_4}\limits^{+6}$.The change in the oxidation state of Manganese $(Mn)$ per atom is: $7 - 6 = 1$.The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}}$.Here,the n-factor is the change in oxidation state,which is $1$.Therefore,the equivalent weight $= \frac{M}{1} = M$.

$M$ is the molecular weight of $KMnO_4$. The equivalent weight of $KMnO_4$ when it is converted into $K_2MnO_4$ is

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