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 × 10 4 kPa for O 2 . Density of water with dissolved oxygen = 1 . 0 kg dm - 3 ]

Step-by-step solution

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