The air is a mixture of a number of gases. The maojr components are oxygen and nitrogen with approximate proportion of `20%` is to `79%` by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry's law constants for oxygen and nitrogen at 298 K are `3.30xx10^(7)` mm and `6.51xx10^(7)` mm respectively, calculate the composition of these gases in water.

Explanation

Percentage of oxygen `(O_(2))` in air `= 20%`
Percentage of nitrogen `(N_(2))` in air `= 79%`
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, `(10 xx 760) mm Hg = 7600 mm Hg`
Partial pressure of oxygen, `p_(0_(2)) =(20)/(100) xx 7600 mm Hg = 1520 mm Hg`
Partial pressure of nitrogen, `p_(N_(2)) = (79)/(100) xx 7600 mmm Hg = 6004 mm` of `Hg`
Now, according to Henrt's law :
`p = K_(H)x`
For oxygen:
`p_(0_(2)) = k_(H).X_(0_(2))`
`rArr x_(0_(2)) =(p_(0_(2)))/(K_(H)) = (1520 mm of Hg)/(3.30 xx 10^(7) mm of Hg)` (Given `K_(H) = 3.30 xx 10^(7) mm of Hg = 4.16 xx 10^(-5))`
For nitrogen,
`p_(N_(2)) = K_(H)X_(N_(2))`
`rArr X_(N_(2)) = (P_(N_(1)))/(K_(H)) = (6004 mm Hg)/(6.51 xx 10^(7)mm Hg) = 9.22 xx 10^(-5)`
Hence, the mole fractions of oxygen and nitrogen in water are `4.61 xx 10^(-5)` and `9.22 xx 10^(-5)` respectively.