QJEEM23Ch44f3: JEE Main 2023 Chemistry - Class 11 JEEM23Ch44f3 Solution: 1   g of a carbonate ( M 2 CO 3 ) on treatment with excess …

When the carbonate reacts with HCl, it produces CO₂, as shown in the following balanced equation:

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

From the problem, we know that 1 g of the carbonate produces 0.01 mol of CO₂.

To determine the molar mass of the carbonate, we can use the stoichiometry of the reaction:

1 mol of $${\mathrm{M}}_2{\mathrm{CO}}_3$$ produces 1 mol of CO2.

So, 0.01 mol of CO₂ corresponds to 0.01 mol of $${\mathrm{M}}_2{\mathrm{CO}}_3$$.

Now, we can find the molar mass of $${\mathrm{M}}_2{\mathrm{CO}}_3$$:

$$\frac{\text{mass of }{\mathrm{M}}_2{\mathrm{CO}}_3}{\text{moles of }{\mathrm{M}}_2{\mathrm{CO}}_3}=\text{molar mass of }{\mathrm{M}}_2{\mathrm{CO}}_3$$

$$\frac{1\text{ g}}{0.01\text{ mol}}=100{\text{ g mol}}^{-1}$$

So, the molar mass of $${\mathrm{M}}_2{\mathrm{CO}}_3$$ is approximately 100 g/mol.