Question

$28\, g \,KOH$ is required to completely neutralise $CO _{2}$ produced on heating $60 \,g$ of impure $CaCO _{3}$ The percentage purity of $CaCO _{3}$ is approximately (molar masses of $KOH$ and $CaCO _{3}$ are $56$ and $100 \,g \,mol ^{-1}$, respectively )

• A. 41.6
• B. 40
• C. 20.8
• D. 83.3

Answer 1

Answer: (A)

The given relation $CaCO_3 \ce{ ->[{2 KOH}][{\Delta}] }CO_2$ means,

2 moles of KOH will neutralise 1 mole of $CO_2$

Given,

(i) $KOH $ ( used ) $=28\, g$

(ii) Molar mass of $KOH (M)=56\, g$

(iii) Molar mass of $CaCO _{3}=100 \,g$

(iv) Impure $CaCO _{3}=60 \,g$

$\because\, 112\, g (56 \times 2)$ of $KOH$ will neutralise

$=100 \,g$ or $CaCO _{3} $

$ \therefore \,28\, g KOH $ will neutralise $=\frac{100 \times 28}{112} $

$=25\, g $ of $CaCO _{3} $

Also,

$\because \,60\, g$ (impure) of $CaCO _{3}$ has $25 \,g$ pure $CaCO _{3}$.

$\therefore \,100\, g$ (impure) $CaCO _{3}$ has pure $CaCO _{3}$

$=\frac{25 \times 100}{60}$

$=41.6\, g$