Passage:
Solubility of gases in liquids is greatly affected by pressure and temperature. Henry gave the quantitative relationship between pressure and solubility of gas in a solvent. According to Henry’s law, partial pressure of a gas above a liquid is directly proportional to its mole fraction in solution and is expressed as \(P = K_H.x\), where \(K_H\) is Henry’s constant and x is mole fraction of gas. \(K_H\) is a function of nature of gas.
F M Raoult gave a quantitative relationship between partial pressures and mole fractions in the binary solution of volatile liquids. Raoult’s law states that for a binary solution of volatile liquids, the partial pressure of each component in the solution is directly proportional to its mole fraction. Thus for a solution of component \(1\) and \(2\), partial pressure of each component \(P_1 = p_1^0x_1\), where \(p_1^0\) is the vapour pressure of pure component \(1\) at the same temperature. Similarly \(P_2 = p_2^0x_2\)
Henry’s law constant for \(O_2\) in water at \(293\) K is \(46.82\) bar. The mole fraction of oxygen in air is \(0.21\). What mole fraction of oxygen is dissolved in \(1\) litre of water at \(2.026\) bar pressure and \(293\) K?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.009
By Henry’s law, the mole fraction of a gas dissolved in a liquid is proportional to the partial pressure of that gas above the liquid:
\( \displaystyle p = K_H\,x \qquad\Rightarrow\qquad x = \frac{p}{K_H} \)
Air is at a total pressure of 2.026 bar and oxygen makes up 0.21 of it, so the partial pressure of oxygen is
\( \displaystyle p_{\ce{O2}} = 0.21 \times 2.026\ \mathrm{bar} = 0.4255\ \mathrm{bar} \)
\( \displaystyle x_{\ce{O2}} = \frac{0.4255\ \mathrm{bar}}{46.82\ \mathrm{bar}} = 0.0091 \approx 0.009 \)
So the mole fraction of dissolved oxygen is 0.009.