If the solubility of a gas in water is 1.22 g/L at 2.75 atm, what is its solubility (in g/L) at 1.0 atm?

0.44 g/L.

Explanation:

This problem is an application on Henry's law. Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid. Solubility of the gas ∝ partial pressure If we have different solubility at different pressures, we can express Henry's law as: S₁/P₁ = S₂/P₂, S₁ = 1.22 g/L and P₁ = 2.75 atm S₂ = ? g/L and P₂ = 1.0 atm So, The solubility of the gas at 1.0 atm (S₂) = S₁. P₂/P₁ = (1.22 g/L x 1.0 atm) / (2.75 atm) = 0.44 g/L.