Question

The Henry's law constant for the solubility of $N_2$ gas in water at $298\text{K}$ is $1\times {10}^5\text{atm}$. The mole fraction of $N_2$ in air is $0.8$. Then the number of moles of $N_2$ dissolved in $10$ moles of water at $298\text{K}$ and $5\text{atm}$ pressure is:

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

Answer 1

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

$\text{Given, mole fraction of}{N}_2\left({\chi }_{N_2}\right)\text{in air}=0.8$
So according
We are considering the pressure to be $5\text{atm}$ here.
So, partial pressure of $N_2$
$=p_{N_2}={\chi }_{N_2}\times P_{\text{total}}$
$=0.8\times 5=4\text{atm}$
Now, according to Henry's law the partial pressure of the gas in the vapour phase is proportional to the mole fraction of the gas in the solution.
$\therefore p_{N_2}=K_H\times x_{N_2}$
Here, $K_H$ is Henry's law constant
Given, $K_H=1\times {10}^5\text{atm}$
$4=1\times {10}^5\times \frac{n_{N_2}}{n_{N_2}+n_{H_{2}O}}$
$\Rightarrow \frac{n_{N_2}+n_{H_{2}O}}{n_{N_2}}=\frac{1\times {10}^5}{4}$
$\Rightarrow \frac{n_{N_2}+10}{n_{N_2}}=0.25\times {10}^5$
Number of moles of $N_2$ is negligible as compared to number of moles of water.
$\Rightarrow n_{N_2}=4\times {10}^{-4}$
So, the number of moles of $N_2$ dissolved in $10$ moles of water at $298\text{K}$ and $5\text{atm}$ pressure is $4\times {10}^{-4}$