Answer:
X(O₂) = 0.323
X(N₂) = 0.677
Explanation:
We have the partial pressures of oxygen (O₂) and nitrogen (N₂):
P(O₂) = 0.20 atm
P(N₂) = 0.80 atm
In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.
s(O₂) = 1.3 x 10⁻³ mol/(L atm)
s(N₂) = 6.8 x 10⁻⁴mol/(L atm)
So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):
C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L
C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L
In 1 liter of water, we have the following number of moles (n):
n(O₂) = 2.6 x 10⁻⁴ mol
n(N₂) = 5.44 x 10⁻⁴ mol
Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:
nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol
Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:
X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323
X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677
