The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is __________ × 10⁻⁵ mol dm⁻³. (Round off to the Nearest Integer).
[Given : Henry's law constant = K H = 8.0 × 10⁴ kPa for O₂. Density of water with dissolved oxygen = 1.0 kg dm⁻³ ]

Solution

Chemistry ma’am · 4 steps

Henry's law gives you a mole fraction, so turn it into moles using the moles of water in one litre.

1. Henry's law. Henry's law links the partial pressure of a gas to its mole fraction in the solution: p = K_H x. The constant here is in kPa, so the result is a mole fraction, not a molarity.

   x_(O₂) = p/K_H = (20 kPa)/(8.0 × 10⁴ kPa) = 2.5 × 10⁻⁴

2. Moles of water. With density 1.0 kg dm⁻³, one dm³ of solution has a mass of about 1000 g, almost all water. That is 1000/18 = 55.56 mol of water.
3. Moles of oxygen. The solution is very dilute, so x_(O₂) ≈ n_(O₂)/n_(H₂O). This gives n_(O₂) = 2.5 × 10⁻⁴ × 55.56 = 1.389 × 10⁻² mol in each dm³.
4. So. The molar solubility is 1389 × 10⁻⁵ mol dm⁻³, so the answer is 1389.

The common slip: 25. Reporting the mole fraction 2.5 × 10⁻⁴ as the molarity gives 25, not 1389. You still have to multiply by 55.56 mol of water per dm³.

The idea to keep. When K_H is given in pressure units, p/K_H is a mole fraction. To get molarity, multiply it by the moles of solvent in one litre, which is 55.56 mol for water.

Answer from the NTA final answer key. Solution by PrepWiser.