The Henry's law constant for a gas in water at 298 K is 2 10^5 bar. If the partial pressure of the gas is 2 bar, what is the molality of the gas in the aqueous solution?

Step-by-step solution

According to Henry's law, the partial pressure of a gas ( p ) is related to its mole fraction ( x ) in solution by the equation: p = K_H x Given: p = 2 bar K_H = 2 10^5 bar Substituting the values: 2 = 2 10^5 x x = 10⁻⁵ The mole fraction of the gas is very small, so the solution is highly dilute. For an aqueous solution, the number of moles of water in 1 kg of solvent is: n_ H₂O = 1000 g 18 g mol ⁻¹ = 55.55 mol For a dilute solution, the mole fraction x n_ gas n_ H₂O n_ gas = x n_ H₂O = 10⁻⁵ 55.55 = 5.55 10⁻⁴ mol S