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?

0.009

The correct answer is option 4. 0.009.

Partial pressure of \(O_2 = 2.026 × 0.21 =0.425\)

Now, applying Henry's law

\(p_{O_2} = K_H × X_{O_2}\)

\(∴ X_{O_2} = \frac{p_{O_2}}{K_H} = \frac{0.425}{46.82} = 0.009\)