Answer:
The correct answer is 0.00080 gram per liter.
Step-by-step explanation:
Based on the given information, the solubility of water is 0.00380 gram per liter, the temperature mentioned is 25 degree C, the partial pressure of oxygen gas is 760 torr, and the mole fraction of oxygen is 0.210. There is a need to determine the solubility of oxygen in water.
Based on Henry's law,
Solubility of oxygen gas = Henry's constant × partial pressure of oxygen gas
Henry's constant, K = solubility of oxygen gas / partial pressure of oxygen gas
= 0.00380 g/L × 1 mol/32 grams / 760 torr × 1 atm/760 torr
= 0.00012 mol/L/atm
= 0.00012 M/atm
Now the partial pressure of the oxygen gas = mole fraction of oxygen × atmospheric pressure
= 0.210 × 1 atm
= 0.210 atm
Now putting the values in Henry's law equation we get,
Solubility of oxygen gas = 0.00012 mol/L/atm × 0.210 at,
= 0.000025 mol/L × 32 gram/mol
= 0.00080 gram per liter