Molecular weight of Mohr's salt is $392$ . Its equivalent weight when it is oxidised by ${\mathrm{KMnO}}_4$ in acidic medium is

We know that formula for Mohr's salt is ${\mathrm{FeSO}}_4\cdot {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_{4}6{\mathrm{H}}_2\mathrm{O}$

And the reaction between Mohr's salt and acidified ${\mathrm{KMnO}}_4$ is

$$\begin{gathered} 2{\mathrm{KMnO}}_4+10{\mathrm{FeSO}}_4\cdot {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_{4}6{\mathrm{H}}_2\mathrm{O}+8{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4 \\ +10{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4+2{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3+8{\mathrm{H}}_2\mathrm{O} \end{gathered}$$ 

Here, the Net valence=1

And, we know that equivalent weight$=\frac{\mathrm{Molecular} \mathrm{Weight}}{\mathrm{Net} \mathrm{positive} \mathrm{valence}}$

So, the equivalent weight Mohr's salt$=\frac{392}{1}$

Hence, option(a) is the answer.