$28 \mathrm{~g} \mathrm{KOH}$ is required to completely neutralise $\mathrm{CO}_2$ produced on heating $60…
$28 \mathrm{~g} \mathrm{KOH}$ is required to completely neutralise $\mathrm{CO}_2$ produced on heating $60 \mathrm{~g}$ of impure $\mathrm{CaCO}_3$. The percentage purity of $\mathrm{CaCO}_3$ is approximately (molar masses of $\mathrm{KOH}$ and $\mathrm{CaCO}_3$ are 56 and $100 \mathrm{~g} \mathrm{~mol}^{-1}$, respectively)
41.6
40
20.8
83.3
Solution
The given relation $\mathrm{CaCO}_3 \xrightarrow[\Delta]{\stackrel{2 \mathrm{KOH}}{\Delta}} \mathrm{CO}_2$ means, 2 moles of $\mathrm{KOH}$ will neutralise 1 mole of $\mathrm{CO}_2$. Given,
(i) $\mathrm{KOH}$ (used) $=28 \mathrm{~g}$
(ii) Molar mass of $\mathrm{KOH}(M)=56 \mathrm{~g}$
(iii) Molar mass of $\mathrm{CaCO}_3=100 \mathrm{~g}$
(iv) Impure $\mathrm{CaCO}_3=60 \mathrm{~g}$
$\because 112 \mathrm{~g}(56 \times 2)$ of $\mathrm{KOH}$ will neutralise
$$
=100 \mathrm{~g} \text { or } \mathrm{CaCO}_3 \text {. }
$$
$\therefore 28 \mathrm{~g} \mathrm{KOH}$ will neutralise $=\frac{100 \times 28}{112}$
$$
=25 \mathrm{~g} \text { of } \mathrm{CaCO}_3 \text {. }
$$
Also,
$\because 60 \mathrm{~g}$ (impure) of $\mathrm{CaCO}_3$ has $25 \mathrm{~g}$ pure $\mathrm{CaCO}_3$.
$\therefore 100 \mathrm{~g}$ (impure) $\mathrm{CaCO}_3$ has pure $\mathrm{CaCO}_3$
$$
=\frac{25 \times 100}{60}=41.6 \mathrm{~g}
$$
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