Some Basic Concepts of Chemistry 1 Question 13
13. $1 \mathrm{~g}$ of a carbonate $\left(M_{2} \mathrm{CO}{3}\right)$ on treatment with excess $\mathrm{HCl}$ produces 0.01186 mole of $\mathrm{CO}{2}$. The molar mass of $M_{2} \mathrm{CO}_{3}$ in $\mathrm{g} \mathrm{mol}^{-1}$ is
(2017 JEE Main)
(a) 1186
(b) 84.3
(c) 118.6
(d) 11.86
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Solution:
- $\underset{1 \mathrm{~g}}{M_{2} \mathrm{CO}{3}}+2 \mathrm{HCl} \longrightarrow 2 \mathrm{M} \mathrm{Cl}+\underset{\substack{0.01186 \ \text { mole }}}{\mathrm{H}{2} \mathrm{O}}+\underset{2}{\mathrm{CO}_{2}}$
Number of moles of $M_{2} \mathrm{CO}{3}$ reacted $=$ Number of moles of $\mathrm{CO}{2}$ evolved
$$ \begin{aligned} & \frac{1}{M}=0.01186 \quad\left[M=\text { molar mass of } M_{2} \mathrm{CO}_{3}\right] \ & M=\frac{1}{0.01186}=84.3 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned} $$
