At 298 K, the mole percentage of N2(g) in air is 80\%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of N2(g) in water at 298 K? (KH for N2 is 6.5 107 mmHg)
- 9.35 10-5
- 9.35 105
- 1.23 10-7
- 1.17 10-4
Answer
(1) 9.35 10-5
Solution
The partial pressure of N2 over water is
pN2 = 0.80 10 atm = 8 atm
pN2 = 8 760 = 6080 mmHg
By Henry's law, pN2 = KH xN2, so
xN2 = 60806.5 107 = 9.35 10-5
Answer: (1)