Question

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

• A. $5\times {10}^{-4}mol$
• B. $3\times {10}^{-5}mol$
• C. $4\times {10}^{-4}mol$
• D. $4\times {10}^{-5}mol$

Answer 1

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

Given,
total pressure is, $P_t=5\text{atm}$ .
${\chi }_{N_2}=0.8K_H=1\times {10}^{-5}\text{atm}$
The partial pressure of nitrogen according to dalton's law is:
${\chi }_{N_2}\times P_t\Rightarrow 0.8\times 5=4\text{atm}$

According Henry's law, $P=K_H\times X$
Substituting values in the above expression, we get
${\chi }_{N_2}=\frac{4\text{atm}}{1\times {10}^{5}atm}=4\times {10}^{-5}$

${\chi }_{N_2}=\frac{n_{N_2}}{n_{N_2}+n_{H_{2}O}}=4\times {10}^{-5}\Rightarrow 4\times {10}^{-5}=\frac{n_{N_2}}{n_{N_2}+10}$
Solving
$\Rightarrow 4\times {10}^{-5}\times n_{N_2}+4\times {10}^{-4}=n_{N_2}$

$\Rightarrow n_{N_2}(1-(4\times {10}^{-5}\left)\right)=4\times {10}^{-4}$

$\Rightarrow n_{N_2}=\frac{4\times {10}^{-4}}{1-(4\times {10}^{-5})}$

Approximating

$\Rightarrow 1-(4\times {10}^{-5})\approx 1$
$\Rightarrow n_{N_2}=4\times {10}^{-4}\text{mol}$