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 $= 1.0 kg {dm}^{- 3}$ ]

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 $= 1.0 kg {dm}^{- 3}$ ]

Aicharya · Accepted Answer

P = K H &#183; x or, 20 &#215; 10 3 = 8 &#215; 10 4 &#215; 10 3 &#215; n O 2 n O 2 + n water&#160; or, 1 4000 = n O 2 n O 2 + n water&#160; = n O 2 n water Means 1 mole water = 18 &#160; gm = 18 &#160; ml dissolves 1 4000 moles O 2 . Hence, molar solubility = 1 4000 18 &#215; 1000 = 1 72 &#160; mol &#160; dm - 3 = 1388 . 89 &#215; 10 - 5 &#160; mol &#160; dm - 3 &#8776; 1389 &#160; &#215; 10 - 5 &#160; mol &#160; dm - 3

The oxygen dissolved in water exerts a partial pressure of 20   kPa in the vapour above water. The…

Solution

The oxygen dissolved in water exerts a partial pressure of 20 &#160; kPa in the vapour above water. The…

Solution

The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The…