MCQ
$1\,g$ of a carbonate $\left( M _2 CO _3\right)$ on treatment with excess $HCl$ produces $0.01 mol$ of $CO _2$ The molar mass of $M _2 CO _3$ is $..........\,g\,mol ^{-1}$. (Nearest integer)
- A$200$
- B$300$
- C$50$
- ✓$100$
✓
Answer
Correct option: D.
$100$
d
$\underset{1\,gm }{ M _2 CO _3}+\underset{\text { Excess }}{2 HCl } \rightarrow \underset{0.02 \text { mole }}{2 MCl }+ H _2 O +\underset{0.01 \text { mole }}{ CO _2}$
$\underset{1\,gm }{ M _2 CO _3}+\underset{\text { Excess }}{2 HCl } \rightarrow \underset{0.02 \text { mole }}{2 MCl }+ H _2 O +\underset{0.01 \text { mole }}{ CO _2}$
From principle of atomic conservation of carbon atom,
Mole of $M _2 CO _3 \times 1=$ Mole of $CO _2 \times 1$
$\frac{1 gm }{\text { molar mass of } M _2 CO _3}=0.01 \times 1$
$\text { Molar mass of } M _2 CO _3=100\,gm / mole$
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