5 grams of sample of $CaC{O}_3$gives $1.98 gm$ $C{O}_{2 }$on heating. Then

$\underset{100 gm}{{\mathrm{CaCO}}_3}\to \mathrm{Cao}+\underset{44 gm}{{\mathrm{CO}}_2}$

$\text{ so }1.98\mathrm{gm}{\mathrm{CO}}_2\text{ given by }\frac{1.98}{44}\times 100=4.5$

$gms CaC{O}_3 Z$

$5 gm$ sample contain $4.5 gm$${\mathrm{CaCO}}_3$

$\mathrm{\%}\text{ of purity }=\frac{4.5}{5}\times 100=90\mathrm{\%}\text{ pure }$