Henry’s law constant of oxygen is 1.4 × 10–3 mol L–1. atm–1 at 298 K. How much of oxygen is dissolved in 100 mL at 298 K when the partial pressure of oxygen is 0.5 atm? (Kerala CEE)

Given: Henry law constant 3 1 H K = 1.4 10 mol L x Partial vapour pressure of oxygen p = 0.5 atm We are interested in determining the solubility of oxygen gas in water. The solubility of oxygen gas in the water is determined through Henry’s law. It is given as follows, H S = K p x Where S is the solubility of the gas. H 3 3 1 S = K p S = 1.4 10 0.5 S = 0.7 10 mol L x >> x x x The solubility of a gas in the given volume is, 3 1 5 100 S = 0.7 10 mol L 7 10 1000 x x x Now, the number of moles of oxygen gas in solution is calculated as follows, 2 5 2 5 3 1 mole of O 3