a 700.0-ml sample of pure water is allowed to come to equilibrium with pure oxygen gas at a pressure of 755 mmhg.

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19-12-2022
Chemistry

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a 700.0-ml sample of pure water is allowed to come to equilibrium with pure oxygen gas at a pressure of 755 mmhg.

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contexto1212 contexto1212 · Answer 1 · 2022-12-22T17:41:34-05:00

A 700 ml of pure water is allowed to come equilibrium with 0.028896 gram of pure oxygen gas. This is calculated from the expression of Henry's law.

Henry's law states that, At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

C = k P gas

C= is the solubility of oxygen gas in units mole gas/L

k = Henry's law constant = 1.3×10−3M/atm.

P gas. = the partial pressure of oxygen gas is 755 mmHg.

1 mmHg = 0.00131578947 atm.

755 mmHg * 0.00131578947 atm. / 1 mmHg = 0.993 atm.

Now calculate the mass of oxygen gas dissolves in the water

C = k P gas

C = 1.3×10−3M/atm. * 0.993 atm.

C = 1.29×10−3M

C = 1.29×10−3M or mole/L

we have 700 mL or 0.700 L of water, we have 9.03 × 10−4 mole of O2.

Now we have to convert moles into grams.

mass of 1 mole of O2 = 32 g

9.03 ×10−4 mole of O2 * 32 g/ mole =0.028896 g

To learn more about Henry's Law please visit:

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