Ask Question
12 September, 13:29

At 298 K, the Henry's law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen.

1. At 298 K, what is the solubility of oxygen in water exposed to air at 1.00 atm?

+5
Answers (1)
  1. 12 September, 17:02
    0
    0.000273 M

    Explanation:

    Henry's states that at constant temperature the amount of a gas that dissolves in a liquid is directly proportional to the partial pressure in of that gas in equilibrium with that liquid.

    Pressure of Oxygen = mole fraction of Oxygen * 1.00 atm

    Mole fraction Oxygen = 21/100 * 1.00atm = 0.21 atm

    Molar solubility of Oxygen = KH * PO2 = 0.0013 * 0.21 = 0.000273 M
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “At 298 K, the Henry's law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen. 1. At 298 K, what is the solubility of oxygen in water ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers