Q. When 10 g of 90% pure lime stone is heated completely, the volume (in litres) of ${CO}_{2}$ is liberated at STP is
A) 22.4
B) 2.24
C) 20.16
D) 2.016
Explanation
(d)
Amount of pure lime stone $\left( CaCO_{3} \right)is\ 10\ g\ of\ 90\%\ sample$
$= \frac{90}{100} \times 10 = 9\ g$
$CaCO_{3}\overset{\ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}\ \ \ \ \ \ \ \ \ \ \ \ }{\rightarrow}CaO + CO_{2}$
100 g 22.4 L
100 g of lime stone gives 22.4 L of $CO_{2}$ at STP
$\therefore$ 9 g of lime stone will give
$= \frac{22.4}{100} \times 9 = 2.016\ L\ CO_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.