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

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

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.