The sulphate of a metal M contains 9.87% of $\mathrm{M}.$ This sulphate is isomorphous with ${\mathrm{ZnSO}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}$. The atomic weight of $\mathrm{M}$ is 

As the given sulphate is isomorphous with ${\mathrm{ZnSO}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}$ its formula would be

${\mathrm{MSO}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}\cdot \mathrm{m}$ is the atomic weight of $\mathrm{M},$ Molecular weight of 

${\mathrm{MSO}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}=\mathrm{m}+32+64+126=\mathrm{m}+222$

Hence $\%$ of $\mathrm{M}$ $=\frac{\mathrm{m}}{\mathrm{m}+222}\times 100=9.87$ (given) or

$\begin{array}{l}100\mathrm{m}=9.87\mathrm{m}+222\times 9.87\text{ or }90.13\mathrm{m}=222\times 9.87\\ \text{ or }\mathrm{m}=\frac{222\times 9.87}{90.13}=24.3\end{array}$