The equivalent weight of a substance is given by the formula:

$$\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n}}$$

where n is the change in oxidation number per molecule of the reactant (the valence change).

To find the equivalent weight of $ KMnO_4 $ when it is converted to $ K_2MnO_4 $, we first need to establish the oxidation states of manganese (Mn) in both compounds.

• In $ KMnO_4 $, Mn has an oxidation state of +7.


• In $ K_2MnO_4 $, Mn has an oxidation state of +6.

The change in oxidation state per $ KMnO_4 $ molecule is from +7 to +6, which is a decrease of 1. Therefore, $ n = 1 $ in this case.

Substituting this value into the formula gives us:

$$\text{Equivalent weight} = \frac{\text{Molecular weight (M)}}{1} = M$$

Thus, the correct option is Option A: M.