Air is enclosed over water in a closed vessel at a total pressure of 5.0 atm . If the mole fraction of oxygen in the air is 0.20 and the Henry's law constant ( K_H ) for oxygen in water at the given temperature is 4 10^4 atm , what is the mole fraction of oxygen dissolved in the water?

Step-by-step solution

According to Dalton's law of partial pressures, the partial pressure of oxygen ( p_ O ₂ ) in the gaseous phase is: p_ O ₂ = Total pressure Mole fraction of O ₂ in air p_ O ₂ = 5.0 atm 0.20 = 1.0 atm According to Henry's law, the partial pressure of a gas is directly proportional to its mole fraction ( x ) in the solution: p_ O ₂ = K_H x_ O ₂ 1.0 atm = (4 10^4 atm ) x_ O ₂ x_ O ₂ = 1.0 4 10^4 = 0.25 10⁻⁴ = 2.5 10⁻⁵ Answer: 2.5 10⁻⁵