A) 56
B) 158
C) 278
D) 392
Correct Answer: D
Solution :
The molecular weight of Mohrs salt \[(FeS{{O}_{4}}\cdot {{(N{{H}_{4}})}_{2}}S{{O}_{4}}\cdot 6{{H}_{2}}O)\] = 56+2 (32 + 64) + 2 (14 + 4) + 6 (18) = 56 +192 + 36 +108 = 392 \[\therefore \] Equivalent weight of reductant \[=\frac{\text{Mol}\text{.wt}\text{.}}{\text{increase}\,\text{in}\,\text{O}\text{.N}\text{.}}=\frac{392}{1}=392\] (\[\because \]\[F{{e}^{2+}}\] changes into \[F{{e}^{3+}}\]in its reactions)
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