Question

A $821\text{mL}{N}_2\left(g\right)$ was collected over liquid water at $300\text{K}$ and 1 atm. If the vapour pressure of $H_{2}O$ is 30 torr then number of moles of $N_2\left(g\right)$ in the moist gas mixture is:

• A. $0.39$
• B. $0.032$
• C. $0.96$
• D. $0.0013$

Answer 1

The correct option is B $0.032$

Given : Volume, V = 821 mL = 0.821 L
Temperature, T = 300 K
Total pressure, $P_{total}=1\text{atm}=760\text{torr}$
Vapour pressure of $H_{2}O$, $P_{H_{2}O}=30\text{torr}$
$P_{total}=P_{N_2}+P_{H_{2}O}$
$760\text{torr}=P_{N_2}+30\text{torr}$
$P_{N_2}=730\text{torr}=\frac{730}{760}\text{atm}$
Using the ideal gas equation, moles of $N_2\left(g\right)$ in moist gas mixture is:
$n_{N_2}=\frac{P_{N_2}V}{RT}$

$n_{N_2}=\frac{\left(\frac{730}{760}\right)\times 0.821}{0.0821\times 300}=0.032$
Hence, the correct answer is option (b).