What will be the ratio of mole fractions of O₂ and N₂ dissolved in water at 25˚C? (Given that Henry’s law constant (in torr) for O₂ and N₂ are 3.3 × 10⁷ and 6.5 × 10⁷ respectively

Question

What will be the ratio of mole fractions of O₂ and N₂ dissolved in water at 25˚C? (Given that Henry’s law constant (in torr) for O₂ and N₂ are 3.3 × 10⁷ and 6.5 × 10⁷ respectively at 25˚C; mole fractions of O₂ and N₂ in air are taken as 0.2 and 0.8 respectively)

Vimal Swarnkar · Accepted Answer

(d) ~0.49. x = p/KH. x(O₂)/x(N₂) = (0.2/3.3 × 10⁷)/(0.8/6.5 × 10⁷) = (0.2 × 6.5)/(0.8 × 3.3) = 0.49 → (d). O₂ is about twice as soluble as N₂, so dissolved air is O₂-rich.

Answer

(a) ~ 0.51

Answer

(b) ~0.66

Answer

(c) ~0.25

Geo-Scientist Prelims 2020 Chemistry Q70: What will be the ratio of mole fractions of O₂ and N₂ dissolved in…

QuestionSolutions / Henry's law