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).

Question

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 ______ $\times 10^{- 5} mol {dm}^{- 3} .$

(Round off to the Nearest Integer).

[Given : Henry's law constant $= K_{H} = 8.0 \times 10^{4} kPa$ for $O_{2} .$ Density of water with dissolved oxygen $= 1.0 kg {dm}^{- 3}$ ]

TestHub · Accepted Answer

P = K H · x or, 20 × 10 3 = 8 × 10 4 × 10 3 × n O 2 n O 2 + n water or, 1 4000 = n O 2 n O 2 + n water = n O 2 n water Means 1 mole water = 18 gm = 18 ml dissolves 1 4000 moles O 2 . Hence, molar solubility = 1 4000 18 × 1000 = 1 72 mol dm - 3 = 1388 . 89 × 10 - 5 mol dm - 3 ≈ 1389 × 10 - 5 mol dm - 3

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