If O₂ gas is bubbled through water at 303 K, the number of millimoles of O₂ gas that dissolve in 1 litre of water is __________. (Nearest Integer)
(Given : Henry's Law constant for O₂ at 303 K is 46.82 k bar and partial pressure of O₂=0.920 bar)
(Assume solubility of O₂ in water is too small, nearly negligible)

Solution

Chemistry ma’am · 4 steps

Henry's law gives you a mole fraction. Turn it into moles using the 55.6 mol of water in a litre.

1. Henry's law. Partial pressure equals Henry's constant times the mole fraction of the dissolved gas, p = K_H x. Convert K_H to bar first: 46.82 kbar = 46820 bar.

   x_(O₂) = p_(O₂)/K_H = 0.920/46820 = 1.965 × 10⁻⁵

2. Moles of water. 1 litre of water is about 1000 g, so n_(H₂O) = 1000/18 = 55.56 mol. The solubility is tiny, so x_(O₂) = n_(O₂)/(n_(O₂) + n_(H₂O)) ≈ n_(O₂)/n_(H₂O).
3. Moles of oxygen. n_(O₂) = 1.965 × 10⁻⁵ × 55.56 = 1.09 × 10⁻³ mol, which is 1.09 mmol.
4. So. About 1.09 millimoles of O₂ dissolve in 1 litre of water. To the nearest integer, that is 1.

The common slip: 1092. Leaving K_H as 46.82 without changing kbar to bar makes the answer 1000 times too large, about 1092 mmol instead of 1.

The idea to keep. Henry's law in the form p = K_H x gives a mole fraction, not a concentration. For a dilute gas in 1 L of water, multiply x by 55.6 mol of water to get the moles of gas.

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