Question

The volume occupied by $16g$ of oxygen gas at $STP$ is

• A. 22.4 L
• B. 44.8 L
• C. 11.2 L
• D. 5.6 L

Answer 1

Answer: (C)

$1$ mole of a gas occupy $=22.4L$ volume at $STP$ therefore, $16g{O}_2$ gas $=0.5$ mole of $O_2$ gas will occupy $=\frac{0.5\times 22.4}{1}=11.2L$ at $STP$.