The Henry’s law constant for the solubility of gas in water at 298 K is atm. The mole fraction of in — Solutions and Colligative Properties Chemistry Question
Question
The Henry’s law constant for the solubility of $N_2$ gas in water at 298 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 dissolved in 10 moles of water at 298 K and 5 atm pressure is
Answer: A
💡 Solution & Explanation
$p_{N_2} = y_{N_2} \times P_{total} = 0.8 \times 5 = 4 \text{ atm}$. According to Henry's law: $p_{N_2} = K_H \times x_{N_2}$. $x_{N_2} = 4 / (1.0 \times 10^5) = 4 \times 10^{-5}$. Since $x_{N_2} = n / (n + 10) \approx n / 10$, we have $n / 10 = 4 \times 10^{-5} \Rightarrow n = 4.0 \times 10^{-4} \text{ moles}$. Therefore, correct answer is A.
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