The Henry's law constant () for the solubility of gas in water at is . The mole fraction of in air i — Solutions and Colligative Properties Chemistry Question
Question
The Henry's law constant ($K_H$) 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 air is $0.8$. Calculate 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 pressure.
Answer: A
💡 Solution & Explanation
Partial pressure $P_{N_2} = 0.8 \times 5 = 4\text{ atm}$. By Henry's law, $\chi_{N_2} = \frac{P}{K_H} = \frac{4}{1.0 \times 10^5} = 4 \times 10^{-5}$. In dilute solution, $\chi_{N_2} \approx \frac{n_{N_2}}{n_{H_2O}} \implies \frac{n_{N_2}}{10} = 4 \times 10^{-5} \implies n_{N_2} = 4 \times 10^{-4}$.
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