Question

Calculate the proportion of O₂ and N₂ dissolved in water at 298 K. When air containing 20% O₂ and 80% N₂ by volume is in equilibrium with water at 1 atm pressure. Henry’s law constants for two gases are KH(O₂) = 4.6 x atm and K_H(N₂) 8.5 x 10⁴ atm.

Answer 1

C₁V₁ = C₂V₂ 

6M(V₁) = 0.25M x 500 ml 

V₁ = \(\frac{0.25\times500}{6}\)

V₁ = 20.3 mL