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 _____×10−5mol dm−3[ Given : Henry’s law constant = KH=8.0×104k Pa for O2, Density of water with dissolved oxygen = 1.0 kg dm−3 ]

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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 _____ $\times 10^{- 5} m o l d m^{- 3}$
[ Given : Henry’s law constant = $K_{H} = 8.0 \times 10^{4} k P a f o r O_{2}$ , Density of water with dissolved oxygen = $1 .0 k g {dm}^{- 3}$ ]

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answer is 1389.

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Detailed Solution

$\begin{array}{l} P = K_{H} X_{gas} \\ 20 \times 10^{3} = 8 \times 10^{4} \times 10^{3} \times X_{gas} \\ 20 \times 10^{3} = 8 \times 10^{4} \times 10^{3} \times \frac{n_{O_{2}}}{n_{O_{2}} + n_{water}} \\ \frac{n_{O_{2}}}{n_{O_{2}} + n_{water}} = \frac{1}{4000} \\ \frac{n_{O_{2}}}{n_{water}} = \frac{1}{4000} \end{array}$

Mass present in one mole of water is 18 g which dissolves in 18 mL (density of water is 1.0 g/mL).

Moles dissolve in $\frac{1}{4000}$ is:

$\frac{\frac{1}{4000}}{18} \times 1000 = \frac{1}{72} mol / {dm}^{3} ≅ 1389 mol / {dm}^{3}$

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