MCQ
$1\, gram$ 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 :
- A$1186$
- ✓$84.3$
- C$118.6$
- D$11.86$
✓
Answer
Correct option: B.
$84.3$
b
$\mathrm{M}_{2} \mathrm{CO}_{3}+2 \mathrm{HCl} \rightarrow 2 \mathrm{MCl}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$
$\mathrm{M}_{2} \mathrm{CO}_{3}+2 \mathrm{HCl} \rightarrow 2 \mathrm{MCl}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$
$n=\frac{\text { Mass given }}{\text { Molar mass }}$
$0.01186=\frac{1}{M} \Rightarrow M=84.3\, \mathrm{g\,mol}^{-1}$
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