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A 600.0-mL sample of pure water is allowed to come to equilibrium with pure oxygen gas at a pressure of 920 mmHg. What mass of oxygen gas dissolves in the water (The Henry's law constant for oxygen gas at 25∘C is 1.3×10−3M/atm)?

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Final answer:

To find the mass of oxygen dissolved in water at a given pressure, we apply Henry's law. The calculated mass of oxygen that dissolves in the 600.0-mL sample of water at 920 mmHg pressure and 25°C is approximately 0.0302 g.

Step-by-step explanation:

The question asks us to calculate the mass of oxygen gas that dissolves in water when exposed to a certain pressure. It involves the application of Henry's law, which relates the concentration of a dissolved gas to its partial pressure above the solution. The Henry's law constant for oxygen at 25°C is given, along with the pressure and volume of the water.

First, convert the pressure from mmHg to atm: 920 mmHg × (1 atm / 760 mmHg) = 1.21 atm. Then use Henry's law to calculate the solubility (molarity, M) of oxygen in water: solubility = Henry's law constant × pressure, which equals 1.3×10⁻³ M/atm × 1.21 atm = 1.573×10⁻³ M.

To find the mass of dissolved oxygen, we need to know the moles of oxygen, which we can calculate by multiplying the solubility by the volume of water (in liters). Finally, the moles are converted to grams using the molar mass of oxygen (32 g/mol).

Moles of O₂ = 1.573×10⁻³ M × 0.600 L = 9.438×10⁻´ mol

Mass of O₂ = 9.438×10⁻´ mol × 32 g/mol = 0.0302 g

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