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

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$130.6$

$196$

$78.5$

$392$

answer is A.

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Detailed Solution

formula of Mohr's salt is ${({NH}_{4})}_{2} Fe {({SO}_{4})}_{2}, 6 H_{2} O$ .

Reaction occurring in acidic medium is given as .

$2 {K M n O}_{4} + 10 {F e S O}_{4} \cdot {({N H}_{4})}_{2} {S O}_{4} \cdot 6 H_{2} O + 8 H_{2} {S O}_{4} \to K_{2} {S O}_{4} + 2 {M n S O}_{4} + 10 {({N H}_{4})}_{2} {S O}_{4} + 2 {F e}_{2} {({S O}_{4})}_{3} + 8 H_{2} O$

The change observed in mohr's salt -

${F e S O}_{4} \cdot {({N H}_{4})}_{2} {S O}_{4} \cdot {S H}_{2} O ⟶ {F e}_{2} {({S O}_{4})}_{3}$

The oxidation state of $F e$ charges from $+ 2$ to $+ 3$ . so, for 1 mole of mohr's salt, 1 mole of electrons is lost. Then, the equivalent weight of mohr's salt is

$= \frac{m o l e c u l a r w e i g h t}{n f a c t o r (e l e c t r o n s e x c h a n g e d)} = \frac{392}{1} = 392 g$