Some Basic Concepts of Chemistry - Result Question 75
####13. $1 g$ of a carbonate $\left(M _2 CO _3\right)$ on treatment with excess $HCl$ produces 0.01186 mole of $CO _2$. The molar mass of $M _2 CO _3$ in $g mol^{-1}$ is
(2017 JEE Main)
(a) 1186
(b) 84.3
(c) 118.6
(d) 11.86
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Solution:
- $\underset{1 g}{M _2 CO _3}+2 HCl \longrightarrow 2 M Cl+\underset{\substack{0.01186 \ \text { mole }}}{H _2 O}+\underset{2}{CO _2}$
Number of moles of $M _2 CO _3$ reacted $=$ Number of moles of $CO _2$ evolved
$$ \begin{aligned} & \frac{1}{M}=0.01186 \quad\left[M=\text { molar mass of } M _2 CO _3\right] \ & M=\frac{1}{0.01186}=84.3 g mol^{-1} \end{aligned} $$
