1 gram of a carbonate M$$_{2}$$CO$$_{3}$$ on treatment with excess HCl produces 0.01186 moles of CO$$_{2}$$. The molar mass of M$$_{2}$$CO$$_{3}$$ in g mol$$^{-1}$$ is:

We are told that a sample of mass 1 g of the carbonate $$M_2CO_3$$ reacts completely with excess hydrochloric acid and that this reaction liberates $$0.01186$$ moles of carbon dioxide.

First, let us write the balanced chemical equation for the reaction between the carbonate and hydrochloric acid:

$$M_2CO_3 + 2\,HCl \rightarrow 2\,MCl + H_2O + CO_2$$

From the stoichiometry of this equation we notice that one mole of the carbonate $$M_2CO_3$$ produces one mole of carbon dioxide $$CO_2$$.

Therefore, the number of moles of the carbonate present in the 1 g sample is exactly the same as the number of moles of carbon dioxide formed.

So we have

$$n(M_2CO_3)=n(CO_2)=0.01186\;\text{mol}$$

Now, the definition of molar mass is

$$\text{Molar mass}=\frac{\text{Mass of sample}}{\text{Number of moles}}$$

Substituting the given mass (1 g) and the calculated moles (0.01186 mol) into this formula we get

$$\text{Molar mass of }M_2CO_3=\frac{1\;\text{g}}{0.01186\;\text{mol}}$$

Carrying out the division,

$$\frac{1}{0.01186}=84.3\;\text{g mol}^{-1}$$

Hence, the molar mass of $$M_2CO_3$$ is $$84.3\;\text{g mol}^{-1}$$.

Hence, the correct answer is Option A.