The Molecular weight of Mohr's salt is 392. Its equivalent weight when it is oxidized by $K M n O_{4}$ , in acidic medium is:

392

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196

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130.6

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78.5

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Solution

The correct option is C 392
The formula of Mohr's salt is FeSO $_{4}$ . (NH $_{4}$ ) $_{2}$ SO $_{4}$ 6H $_{2}$ O .

The reaction between Mohr's salt and acidified KMnO $_{4}$ is given as :

2KMnO $_{4}$ +10 FeSO $_{4}$ . (NH $_{4}$ ) $_{2}$ SO $_{4}$ 6H $_{2}$ O + 8H $_{2}$ SO $_{4}$ $\to$ K $_{2}$ SO $_{4}$ +2MnSO $_{4}$ +10(NH $_{4}$ ) $_{2}$ SO $_{4}$ +2Fe $_{2}$ (SO $_{4}$ ) $_{3}$ +8H $_{2}$ O.

Here, the oxidation state of Mn changes from +7 to +2 and

the oxidation state of Fe changes from +2 to +3.

So, the Equivalent weight of Mohr's salt = $\frac{M o l e c u l a r w e i g h t o f M o h r s a l t}{1}$
$= 392 g$

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