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The ratio of solubilities of N2 and O2 gases in water from air at 25 degrees C and 1 atm total pressure is (given that air contains 80% N2 and 20% O2 by volume, and Henry's law constants K_H(N2) = 8.42 * 10⁴ atm and K_H(O2) = 4.46 * 10⁴ atm in terms of mole fraction):

1. Approximately 1: 1
2. Approximately 2: 1
3. Approximately 1: 2
4. Approximately 4: 1

Correct answer: Approximately 2: 1

Solution

By Henry's law, mole fraction of dissolved gas x = p / K_H. x(N2) = 0.80 / 8.42*10⁴; x(O2) = 0.20 / 4.46*10⁴. Ratio x(N2)/x(O2) = (0.80 * 4.46) / (0.20 * 8.42) = 3.568 / 1.684 ≈ 2.12: 1 ≈ 2: 1.

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