$1 \mathrm{g}$ of a carbonate $\left({\mathrm{M}}_2{\mathrm{CO}}_3\right)$ on treatment with excess $\mathrm{HCl}$ produces $0.01 \mathrm{mol}$ of ${\mathrm{CO}}_2$. The molar mass of ${\mathrm{M}}_2{\mathrm{CO}}_3$ is ${\mathrm{gmol}}^{-1}$. (Nearest integer)

From Balanced chemical reaction,

${\mathrm{M}}_2{\mathrm{CO}}_3+2\mathrm{HCl}\to 2\mathrm{MCl}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

Here from this, we get:

$0.01 \mathrm{mol} {\mathrm{CO}}_2\equiv 0.01 \mathrm{mol} {\mathrm{M}}_2{\mathrm{CO}}_3\equiv 1 \mathrm{g} {\mathrm{M}}_2{\mathrm{CO}}_3$

 We know that,

Number of moles = $\frac{\mathrm{Given} \mathrm{mass}}{\mathrm{Molar} \mathrm{mass}}$

$\Rightarrow$Molar mass of ${\mathrm{M}}_2{\mathrm{CO}}_3=\frac{\mathrm{Given} \mathrm{mass} \mathrm{of} {\mathrm{M}}_2{\mathrm{CO}}_3}{\mathrm{Number} \mathrm{of} \mathrm{Moles} \mathrm{of} {\mathrm{M}}_2{\mathrm{CO}}_3}=\frac{1}{0.01}=100 \mathrm{g}/\mathrm{mol}$