Question

The Henry's law constant for the solubility of $N_2$ gas in water at $298\text{K}$ is $1.0\times {10}^5$ atm. The mole fraction of $N_2$ in air is $0.8$. The number of moles of $N_2$ from air that dissolves in $10$ moles of water at $298\text{K}$ and $5$ atm pressure is:

• A. $4.0\times {10}^{-4}\text{mol}$
• B. $4.0\times {10}^{-5}\text{mol}$
• C. $5.0\times {10}^{-4}\text{mol}$
• D. $4.0\times {10}^{-6}\text{mol}$

Answer 1

The correct option is A $4.0\times {10}^{-4}\text{mol}$

According to Henry's law, the partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.
Here, partial pressure of $N_2$ would be (p)$=(0.8\times 5)\text{atm}=4\text{atm}$
Given, Hnery's law constant $\left(K_H\right)=1.0\times {10}^5\text{atm}$
Let . mole fraction of $N_2$ in the solution be ${\chi }_{N_2}$
So, according to Henry's law
$4=1.0\times {10}^5{\chi }_{N_2}$
$\Rightarrow {\chi }_{N_2}=4\times {10}^{-5}$
let moles of $N_2$ in 10 moles of water be $x$
Then,
$4\times {10}^{-5}=\frac{x}{x+10}$
$\Rightarrow 4\times {10}^{-4}=(1-0.00004)x\approx x$