Equivalent weight of $K M n O_{4}$ in acidic medium is $31.6$ . What is corresponding value in neutral medium?

52.66 g / e q u i v a l e n t

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56.36 g / e q u i v a l e n t

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55.66 g / e q u i v a l e n t

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None of these

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Solution

The correct option is A $52.66 g / e q u i v a l e n t$
In neutral medium, the net reaction is $M n O_{4}^{-} (+7) \to M n O_{2} (+4)$
The numbers in the parenthesis represents the oxidation number of Mn.
The change in the oxidation number is $7 - 4 = 3$
$Equivalent mass \frac{Formula mass}{Change in the oxidation number} = \frac{158}{3} = 52.67$

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