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The equivalent weight of Mohr's salt, ${FeSO}_{4} . {({NH}_{4})}_{2} {SO}_{4} . 6 H_{2} O$ in a redox reaction is equal to its

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molecular weight/2

molecular weight/3

molecular weight

molecular weight/6

answer is D.

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Detailed Solution

${Fe}^{+ 2} \to {Fe}^{+ 3}$

Valence factor for ${FeSO}_{4} . {({NH}_{4})}_{2} {SO}_{4} . 6 H_{2} O$ = 1 Equivalent weight = molecular weight

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