Question

The equivalent weight of Mohr's salt, $FeS{O}_4$. $\left(N{H}_4\right)2S{O}_4.6H_{2}O$ is equal to (i) atomic weight, (ii) its molecular weight, (iii) half its molecular weight, (iv) one third its molecular weight.

Answer 1

In Mohr's salt oxidation of $F{e}^{2+}$ to $F{e}^{3+}$ involves only one electron change.

$\therefore$ Equivalent weight= $\frac{\text{Molecular weight}}{1}$

$\text{Molecular weight}=392$