The equivalent weight of ${\mathrm{CuSO}}_4$ when it is converted to ${\mathrm{Cu}}_2{\mathrm{I}}_2 [\mathrm{M}=\mathrm{mol}.\mathrm{wt}]$

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

When

$2{\mathrm{CuSO}}_4\to {\mathrm{Cu}}_2{\mathrm{I}}_2$

The net valence=1

Therefore, the equivalent weight of ${\mathrm{CuSO}}_4$$=\mathrm{M}/1$

Hence, option(a) is the answer.