The Henry's law constant for the solubility of N₂ gas in water at 298 K is 1.0 × 10⁵ atm. The mole fraction of N₂ in air is 0.8. The number of moles of N₂ from air dissolved in 10 moles of water at 29

Question

The Henry's law constant for the solubility of N₂ gas in water at 298 K is 1.0 × 10⁵ atm. The mole fraction of N₂ in air is 0.8. The number of moles of N₂ from air dissolved in 10 moles of water at 298 K and 5 atm pressure is

Prashant Kotian · Accepted Answer

Correct answer: A. P_N₂ = 0.8 × 5 = 4 atm. By Henry's law: x_N₂ = P_N₂/K_H = 4/(1.0 × 10⁵) = 4.0 × 10⁻⁵. n_N₂ = x_N₂ × n_H₂O = 4.0 × 10⁻⁵ × 10 = 4.0 × 10⁻⁴ mol.

💡 Solution & Explanation

Practice 22,000+ questions like this