Henry's Law constants for aqueous solutions at 25°C are 8.20 × 10^-7 molal/mmHg for N₂ and 1.62 × 10^-6 molal/mmHg for O₂. Determine the solubility of oxygen in water under an atmospheric pressure of 760 mmHg, assuming that air is 80% N₂ and 20% O₂. A) 6.2 × 10^-3 m B) 2.5 × 10^-4 m C) 1.3 × 10^-4 m D) 5.0 × 10^-4 m E) 1.2 × 10^-3 m

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Quizplus · Accepted Answer

The partial pressure of O₂ is 20% of 760 mmHg, which is 152 mmHg. Using Henry's Law, solubility = Henry's constant × partial pressure = 1.62 × 10⁻⁶ molal/mmHg × 152 mmHg = 2.5 × 10⁻⁴ m.

Henry's Law Constants for Aqueous Solutions at 25°C Are 8