Question

$4.4gm$ of a gas at STP occupies a volume of $2.24$ lit the gas can be:

• A. $O_2$
• B. $CO$
• C. ${NO}_2$
• D. ${CO}_2$

Answer 1

Answer: (D)

${CO}_2$

At STP,

$2.24litre$ is ocupied by $4.4gm$ of gas

Then, $22.4L$ is occupied by $=\frac{22.4\times 4.4}{2.24}=44g$

Molar mass of $C{O}_2$

$C=12,O=16$

$\Rightarrow 12+2\left(16\right)\Rightarrow 12+32=44gm$

So, the gas will be $44g$ .