Question

$12.5$ g of an impure sample of limestone on heating gives $4.4$ g of carbon dioxide. The percentage purity of $CaC{O}_3$ in the sample is:

• A. $72$%
• B. $75$%
• C. $80$%
• D. $85$%

Answer 1

The correct option is C $80$%

${CaCO}_3\stackrel{\Delta }{=}CaO+{CO}_2\uparrow$

$100gm$ $44gm$

Therefore $1$ mole i.e. $100gm$ of ${CaCO}_3$ (lime stone) give $1$ mole i.e. $44gm$ of ${CO}_2$.

$4.4gm$ of ${CO}_2$ are produced from $\frac{100\times 4.4}{44}gm$ i.e. $10gm$ of ${CaCO}_3$.

$\therefore$ The percentage of purity $=(1-\frac{12.5-10}{12.5})\times 100$% $=80$%

$\therefore$ Correct answer is $C$ $(80$%$)$.