Question

$10g$ of limestone on heating produces $4.2g$ of $CaO$. the percentage purity of $Ca{CO}_3$ in limestone is:

[Atomic mass of $Ca=$ $40$]

• A. $85$
• B. $75$
• C. $95$
• D. $80$

Answer 1

Answer: (B)

$75$

$10g$ of limestone i.e. $CaC{O}_3$ contains= $\frac{10g}{100g/mole}$ moles of $CaC{O}_3=0.1$ moles of $CaC{O}_3$

$CaC{O}_3⟶CaO+C{O}_2$

$1$ mole of $CaC{O}_3$ produce $1$ mole of $CaO$

Thus $0.1$ moles of $CaC{O}_3$ must produce $0.1$ mole of $CaO$

$10g$ of $CaC{O}_3$ must produce $0.1\times 56=5.6g$ of $CaO$

But $CaO$ produce is $4.2g$

Pure product obtained is $4.2g$ from $10g$ of $CaC{O}_3$

Product that obtain along with $1$ m purity from $10g$ of $CaC{O}_3$ is $5.6g$

So, percentage purity= $\frac{\text{mass of pure substance obtained}}{\text{mass of impure substance obtained}}\times 100$

% purity= $\frac{4.2}{5.6}\times 100=75$%