Air contains O2 and N2 in the ratio of 1:4 . Calculate the ratio of so...
Ratio of Solubility of O2 and N2 in Water
Given Information:
- Air contains O2 and N2 in the ratio of 1:4
- Henry's constant for O2 = 3.30×107 Torr
- Henry's constant for N2 = 6.60×107 Torr
Solution:
We know that the mole fraction of O2 and N2 in air is 1:4.
Let the mole fraction of O2 and N2 dissolved in water be x and y respectively.
According to Henry's law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. Therefore, we can write:
x = (Henry's constant for O2) x (Partial pressure of O2 in air)
y = (Henry's constant for N2) x (Partial pressure of N2 in air)
Let the partial pressure of O2 and N2 in air be P and 4P respectively (as they are in the ratio of 1:4).
Then, x = (3.30×107 Torr) x (P/T) and y = (6.60×107 Torr) x (4P/T), where T is the total pressure of air (which is atmospheric pressure).
Now, we can write the ratio of solubility of O2 and N2 as:
x:y = (3.30×107 Torr) x (P/T) : (6.60×107 Torr) x (4P/T)
x:y = 1:2
Conclusion:
The ratio of solubility of O2 and N2 in water is 1:2.