The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with ZnSO4.7H2O. The atomic weight of M is

The correct option is 24.3.

Isomorphism is a phenomenon where two or more compounds have similar chemical formulas, crystalline structures, and shape. Since the sulphate of the metal M is isomorphous with zinc sulphate heptahydrate (${\mathrm{ZnSO}}_4·7H_{2}O$), the formula of the metal sulphate must be analogous to it.
Therefore, the chemical formula of the hydrated metal sulphate is:
${\mathrm{MSO}}_4·7H_{2}O$

Let the atomic weight of the metal M be represented by $A$.
Let us calculate the total molar mass of the compound ${\mathrm{MSO}}_4·7H_{2}O$:
The atomic weights of the constituent elements are:
Sulfur ($S$) = 32
Oxygen ($O$) = 16
Hydrogen ($H$) = 1
Thus, the molar mass of the ${\mathrm{SO}}_4$ group is:
$32+(4\times 16)=32+64=96$
The molar mass of the $7H_{2}O$ molecules is:
$7\times (2\times 1+16)=7\times 18=126$
So, the total molecular weight of ${\mathrm{MSO}}_4·7H_{2}O$ is:
$\mathrm{Molar\; Mass}=A+96+126=A+222$

According to the problem, the metal M constitutes 9.87% of the total mass of the sulphate.
We can set up the percentage composition formula as follows:
$\left(\frac{A}{A+222}\right)\times 100=9.87$

Now, we solve for $A$:
$100A=9.87\times (A+222)$
$100A=9.87A+2191.14$
Subtract $9.87A$ from both sides:
$(100-9.87)A=2191.14$
$90.13A=2191.14$
Divide by 90.13:
$A=\frac{2191.14}{90.13}\approx 24.3$

Thus, the atomic weight of the metal M is approximately 24.3, which corresponds to the element magnesium (Mg).