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 = $1.0 K g d m^{- 3}$ ]

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Detailed Solution

$\begin{array}{l} P = K_{H} X_{g a s} \\ 20 = 8 \times 10^{4} X_{g a s} \\ x_{g a s} = 2.5 \times 10^{- 4} \\ X_{g a s} = \frac{m}{m + \frac{1000}{18}} \end{array}$

Neglecting amount of gas dissolved when compared with moles of water

$\begin{aligned} m = 0.013888 = 1388.8 \times 10^{- 5} \\ = 1389 \times 10^{- 5} \\ Molarity(M) ≃ Molality(m) \end{aligned}$

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