$100 \mathrm{g}$ or one mole $22400 \mathrm{ml}$
$100$ $g$ of $\mathrm{CaCO}_{3}$ gives $22400 \mathrm{ml}$ of carbon dioxide.
$10 \mathrm{g}$ of $\mathrm{CaCO}_{3}$ gives $=\frac{22400}{100} \times 10=2240 \mathrm{ml}$ of carbon dioxide.
If the metal is $100 \%$ pure, $2240$ ml of carbon dioxide is produced.
$1120 \mathrm{ml}$ of carbon dioxide is produced in this case.
So the $\%$ purity $=\frac{100}{2240} \times 1120=50 \%$
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$Ph - C \equiv C - Ph \to $ 
