In $K M n O_{4}$ titration in acidic meadium, equivalent mass of $K M n O_{4}$ is:

M

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\frac{M}{5}

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\frac{M}{2}

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\frac{M}{3}

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Solution

The correct option is B $\frac{M}{5}$
In acidic medium $K M n O_{4}$ undergoes five electron change.
$+ 7 M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

Equivalent Mass = $\frac{Molecular Mass}{No. of electron gain or lost} = \frac{M}{5}$

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Similar questions

K M n O_{4}

M

List - I	List - II
A) Equivalent weight of $K M n O_{4}$ in acidic medium	1) M/6
B) $K M n O_{4}$ in neutral medium	2) M/5
C) $K M n O_{4}$ in basic medium	3) M/3
D) Phosphorous acid	4) M/2
	5) M/1

K M n O_{4}

K M n O_{4}

K M n O_{4}

K M n O_{4}

M n O_{4}^{-}

K M n O_{4} \to M n^{2 +}

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