The Henry's law constant for the solubility of gas in water at is . The mole fraction of in the atmo — Solutions and Colligative Properties Chemistry Question
Question
The Henry's law constant for the solubility of $N_2$ gas in water at $298\text{ K}$ is $1.0 \times 10^5\text{ atm}$. The mole fraction of $N_2$ in the atmosphere is $0.8$. If the exact number of moles of $N_2$ from air dissolved in $10\text{ moles}$ of water at $298\text{ K}$ and $5\text{ atm}$ total atmospheric pressure is expressed as $Y \times 10^{-4}$, what is the value of $Y$?
💡 Solution & Explanation
Partial pressure of $N_2 = y_{N_2} \times P_{total} = 0.8 \times 5 = 4\text{ atm}$. By Henry's law, $P_{N_2} = K_H \times \chi_{N_2}$. So, $\chi_{N_2} = \frac{4}{1.0 \times 10^5} = 4 \times 10^{-5}$. The mole fraction in solution $\chi_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}} \approx \frac{n_{N_2}}{10}$. Therefore, $\frac{n_{N_2}}{10} = 4 \times 10^{-5} \implies n_{N_2} = 4 \times 10^{-4}$. Hence, $Y = 4$.