In the presence of dilute sulphuric acid, the equivalent weight of potassium permanganate is:

\frac{1}{5}

of its molecular weight

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\frac{1}{6}

of its molecular weight

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\frac{1}{10}

of its molecular weight

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\frac{1}{2}

of its molecular weight

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Solution

The correct option is A $\frac{1}{5}$ of its molecular weight
$4 K M n O_{4} + 6 H_{2} S O_{4} \to 4 M n S O_{4} + 2 K_{2} S O_{4} + 5 O_{2} + 6 H_{2} O$

$M n^{7 +} + 5 e^{-} \to M n^{2 +}$

The number of electrons gained by $K M n O_{4}$ in acidic solution is $5$ .

Therefore, $E q . w t . = \frac{M o l w t .}{5}$

Hence, option A is correct.

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