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A gas mixture contains 20% by volume of O2 and 80% by volume of N2. Using Henry's law constants K_H(N2) = 2 * 10⁴ atm and K_H(O2) = 1 * 10⁴ atm, find the ratio of moles of N2 to moles of O2 dissolved in water when the mixture is in contact with water at the given partial pressures.

1. 8: 1
2. 1: 8
3. 2: 1
4. 1: 2

Correct answer: 2: 1

Solution

Henry's law states x = P / K_H. For N2: x(N2) = 0.80 / (2*10⁴) = 4*10⁻⁵. For O2: x(O2) = 0.20 / (1*10⁴) = 2*10⁻⁵. So moles ratio N2: O2 = 4*10⁻⁵: 2*10⁻⁵ = 2: 1. But the question asks the ratio of dissolved N2 to O2, which is 2:1. Wait — checking options: the answer that matches N2:O2 = 2:1 is option C. However reconsidering: x(N2)/x(O2) = [0.80/20000] / [0.20/10000] = [4*10⁻⁵] / [2*10⁻⁵] = 2. So N2 dissolved: O2 dissolved = 2:1.

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