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Molecular weight of Mohr's salt is 392. Its equivalent weight when it is oxidised by ${KMnO}_{4}$ in acidic medium is

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392

196

130.6

78.5

answer is A.

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Detailed Solution

Eq .wt = \frac{Mol .wt}{n}

‘n’ for Mohr’s salt in acidic medium $= 1$

$Eq .wt = \frac{M .wt}{1} = \frac{392}{1} = 392$

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