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 6.5 10⁷ ~mm Hg )

Step-by-step solution

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