Explanation

(a)
Mohr’s salt is $\left( NH_{4} \right)_{2}SO_{4}.FeSO_{4}.6H_{2}O$
The equation is
$5Fe^{2 +} + MnO_{4}^{-} + 8H^{+} \rightarrow 5Fe^{3 +} + Mn^{2 +} + 4H_{2}O$
Total change in oxidation number of iron
$= ( + 3) - ( + 2)$
$= + 1$
So, equivalent wt. of Mohr’s salt
$= \frac{Mol.wt.of\ Mohr^{'}s\ salt}{1}$
$= \frac{392}{1}$
$= 392$