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  • The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The moral solubility of oxygen in water is ___________<img alt="\times 10^{-5} \mathrm{~mol} \mathrm{\ dm}^{-3}" src="/latex-image/?%5Ctimes%2010%5E%7B-5%7D%20%5C

The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The moral solubility of oxygen in water is ___________\times 10^{-5} \mathrm{~mol} \mathrm{\ dm}^{-3}. (Round off to the Nearest Integer) [Given : Henry's law constant =\mathrm{K}_{\mathrm{H}}=8.0 \times 10^{4} \mathrm{\ kPa} \text { for } \mathrm{O}_{2}               Density of water with dissolved oxygen =1.0 \mathrm{~kg} \mathrm{\ dm}^{-3}]
Option: 1 25
Option: 2 -
Option: 3 -
Option: 4 -

Answers (1)

best_answer

The effect of pressure on the solubility of the gas in the liquid is given by Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution at a definite temperature.

The solubility is taken as the mass of the gas dissolved per unit volume of the liquid. Thus, if m is the mass of the gas dissolved per unit volume of the solvent and P is the pressure of the gas in equilibrium with the solution, then

\begin{aligned} &m \propto P \\ &m=K P \end{aligned}

where K is the proportionality constant. 

\mathrm{p=\frac{1}{K}m =K_{H} m=K_{H} \times\textup{Solubility}}

So,

\textup{Solubility}=\frac{\mathrm{p}}{\mathrm{K_{H}} }

\text { Solubility }=\frac{20 \times 10^{3}}{8.0 \times 10^{7}}=2.5 \times 10^{-4}

\text { Solubility }=25 \times 10^{-5}

Ans = 25

Posted by

Kuldeep Maurya

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