1 gram of a carbonate $(M_2CO_3)$ on treatment with excess $HCl$ produces $0.01186$ mole of $CO_2$. The molar mass of $M_2CO_3$ in g $mol^{-1}$ is :

Question

A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is:
(Given atomic masses of A=64; B=40; C=32 u)

Answer 1

To find the molar mass of the carbonate M_2CO_3, we need to analyze the reaction between the carbonate and excess hydrochloric acid:

The reaction is as follows:

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

From the reaction, 1 mole of M_2CO_3 produces 1 mole of CO_2.

We are given that 0.01186 moles of CO_2 are produced. Thus, 0.01186 moles of M_2CO_3 reacted.

The mass of M_2CO_3 used is 1 gram. Therefore, we use the formula for molar mass:

\text{Molar Mass of } M_2CO_3 = \frac{\text{Mass}}{\text{Moles}}

Substitute the given values:

\text{Molar Mass of } M_2CO_3 = \frac{1 \text{ gram}}{0.01186 \text{ moles}} \approx 84.3 \text{ g/mol}

Thus, the molar mass of M_2CO_3 is 84.3 \, \text{g/mol}, which matches one of the given options.

In conclusion, the correct answer is 84.3 \, \text{g/mol}.

Answer 2