At 298 K, the mole percentage of N _ 2 ( ~g ) in air is 80 \%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of N _ 2 ( ~g ) in water at 298 K ? ( K _ H for N _ 2 is 6.5 10^ 7 ~mm Hg )

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Correct answer: D. 9.35 10^ -5

Using Henry's Law: Mole fraction = P K_H N₂ mole percentage in air = 80%, total pressure = 10 atm Partial pressure of N₂ = 0.80 × 10 = 8 atm = 8 × 760 = 6080 mmHg K_H for N₂ = 6.5 × 10⁷ mmHg Mole fraction of N₂ = 6080 6.5 10^7 = 6.08 10^3 6.5 10^7 = 0.935 10^ -4 = 9.35 10^ -5

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