If 0.0106 g of a gas dissolves in 0.792 L of water at 0.321 atm, what quantity of this gas (in grams) will dissolve at 5.73 atm?

0.189 g.

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 = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm

S₂ = ? g/L and P₂ = 5.73 atm

So, The solubility of the gas at 5.73 atm (S₂) = S₁. P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.

The quantity in (g) = S₂ x V = (0.239 g/L) (0.792 L) = 0.189 g.