$$1.1$$ g of a carbonate M$$_2$$CO$$_3$$ on treatment with excess HCl produces $$0.01$$ mol of CO$$_2$$. The molar mass of M$$_2$$CO$$_3$$ is _____ g mol$$^{-1}$$. (Nearest integer)

We are given that 1.1 g of a carbonate $$\text{M}_2\text{CO}_3$$ on treatment with excess HCl produces 0.01 mol of $$\text{CO}_2$$.

Write the balanced equation: $$ \text{M}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{MCl} + \text{H}_2\text{O} + \text{CO}_2 $$

Find moles of $$\text{M}_2\text{CO}_3$$. From the balanced equation, 1 mol of $$\text{M}_2\text{CO}_3$$ produces 1 mol of $$\text{CO}_2$$.

Therefore, moles of $$\text{M}_2\text{CO}_3$$ = moles of $$\text{CO}_2$$ = 0.01 mol.

Calculate molar mass: $$ \text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{1.1}{0.01} = 110 \text{ g mol}^{-1} $$

Verification: If molar mass of $$\text{M}_2\text{CO}_3$$ = 110, then $$2M + 12 + 48 = 110$$, giving $$M = 25$$. This corresponds to a monovalent metal with atomic mass 25.

The molar mass of $$\text{M}_2\text{CO}_3$$ is 110 g mol$$^{-1}$$.