Question

The air contains $78$ of $N_2$ & $22\%$ of $O_2$ by volume. The volume occupied by $40gm$ air at ${20}^{o}C$ & $745$ mm Hg pressure approximately:

• A. $34$ lit
• B. $34$ ml
• C. $3.4$ K. lit.
• D. none of these

Answer 1

The correct option is D none of these

Solution:-

Since the air contains $78\%N_2$ and $22\%O_2$ by volume.

Therefore, given weight of air $=40\text{gm}$

Mass of $N_2$ in $40\text{gm}$ of air $=\frac{78}{100}\times 40=31.2\text{gm}$

Mass of $O_2$ in $40\text{gm}$ of air $=\frac{22}{100}\times 40=8.8\text{gm}$

Molar mass of $N_2=28\text{gm}$

Molar mass of $O_2=32\text{gm}$

As we know that,

$\text{No. of moles}=\frac{\text{mass}}{\text{molar mass}}$

$\therefore \text{No. of moles in of}{N}_2\text{in 40 gm of air}=\frac{31.2}{28}=1.114\text{mole}$

$\text{No. of moles in of}{O}_2\text{in 40 gm of air}=\frac{8.8}{32}=0.275\text{mole}$

As we know that volume occupied by 1 mole of a gas is $22.4L$.

Therefore,

Volume occupied by 1.114 moles of $N_2=22.4\times 1.114=24.95L$

Volume occupied by 0.275 moles of $O_2=22.4\times 0.275=6.16L$

$\therefore$ Total volume occupied by $40\text{gm}$ of air $=V_{N_2}+V_{O_2}=(24.95+6.16)L=31.11L$

Hence, the volume occupied by $40\text{gm}$ of air is $31.11L$.