The equivalent weight of Mohr’s salt $(N H_{4})_{2} F e (S O_{4})_{2} . 6 H_{2} O$ is equal to its :

Molecular weight / 2

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Atomic weight / 2

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Molecular weight / 3

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Molecular weight

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Solution

The correct option is D Molecular weight
$F e^{2 +} \to F e^{3 +} + e^{-}$

$F e S O_{4}$ present in Mohr’s salt gets converted into $F e_{2} (S O_{4})_{3}$ . This is a one-electron change.

$E q u i v a l e n t w e i g h t = \frac{M o l e c u l a r w e i g h t}{n - F a c t o r}$

$= M o l e c u l a r w e i g h t .$

Hence, option D is correct.

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(N H_{4}) 2 S O_{4} .6 H_{2} O

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