10 grams of limestone on calcination liberates 1.68 lit  of ${\mathrm{CO}}_2$ at STP. Percentage purity of limestone is

${\mathrm{CaCO}}_3\left(\mathrm{S}\right)\stackrel{ }{\to }\mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

100 g                                                     22.4 lit

10 g                                                        2.24 lit    

10 grams of 100% pure limestone liberates 2.24 lit ${\mathrm{CO}}_2$ of  at STP

10 grams of “X%” pure limestone liberates 1.68 lit of ${\mathrm{CO}}_2$  at STP

Percentage purity  (X)  = $\frac{1.68}{2.24}\times 100$

                                                = 75 %