$K M n O_{4}$ acts as an oxidising agent in the neutral medium and gets reduced to $M n O_{2}$ . The equivalent weight of $K M n O_{4}$ in neutral medium is ______ .

mol.wt/2

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mol.wt/3

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mol.wt/4

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mol.wt/7

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Solution

The correct option is B
mol.wt/3

In neutral medium $M n^{+ 7}$ oxidation state changes into +4 oxidation state, hence equivalent weight of $K M n O_{4} = \frac{M}{3}$

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Similar questions

K M n O_{4}

M n O_{2}

K M n O_{4}

K I O_{3}

2 C r (O H)_{3} + 4 O H + K I O_{3} \to 2 C r O_{4}^{2 -} + 5 H_{2} O + K I

K_{2} {C r}_{2} O_{7} (M o l . w t . = M)

K M n O_{4}

K M n O_{4}

K I O_{3}

2 C r (O H)_{3} + 4 O H + K I O_{3} \to 2 C r O_{4}^{2 -} + 5 H_{2} O + K I

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