Henry's law constant for N2 and O2 is 1.6 *10^-6and 8.77*10^-7 molal/mm Hg at 298k.Air has 78 mole percentage N2 and 21 mole percentage O2.at atmospheric pressure the ratio of solubility of N2 and O2 in water is?
To find the ratio of the solubility of N2 and O2 in water, we can use Henry's law.
Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Mathematically, it can be expressed as:
C = k * P
where:
C is the solubility of the gas in the liquid
k is the Henry's law constant for the gas
P is the partial pressure of the gas
Now, let's apply this to N2 and O2 separately.
For N2:
C(N2) = k(N2) * P(N2)
For O2:
C(O2) = k(O2) * P(O2)
Given that the mole percentage of N2 in air is 78% and O2 is 21%, we can assume that the partial pressure of N2 is 78% of the total pressure and the partial pressure of O2 is 21% of the total pressure.
Now, let's calculate the partial pressures of N2 and O2:
Partial pressure of N2 (P(N2)) = 0.78 * P
Partial pressure of O2 (P(O2)) = 0.21 * P
Substituting these values into the equations:
C(N2) = k(N2) * (0.78 * P)
C(O2) = k(O2) * (0.21 * P)
Now, we can calculate the ratio of their solubilities:
Ratio = C(N2) / C(O2)
= (k(N2) * 0.78 * P) / (k(O2) * 0.21 * P)
= (k(N2) / k(O2)) * (0.78 / 0.21)
Substituting the given values of the Henry's law constants:
Ratio = (1.6 * 10^-6) / (8.77 * 10^-7) * (0.78 / 0.21)
This simplifies to:
Ratio = 6.01
Therefore, the ratio of the solubility of N2 to O2 in water is approximately 6.01.