Henry's law constant for oxygen and nitrogen dissolved in water at 298 K are `2.0 xx 10^(9)` Pa and` 5.0 xx 10^(9)` Pa , respectively . A sample of water at a temperature just above 273 K was equilibrated with air (20% oxygen and 80% nitrogen ) at 1 atm. The dissolved gas was separated from a sample of this water and the dried. Determine the composition of this gas.

Explanation

The given values are:
` K_(H(O2))= 2 xx 10^(9) Pa, K_(H(N2)) = 5 xx 10^(9)` Pa
Partial pressure of oxygen,
`P_(O_2)` = 0.2 atm = 20265 Pa
Partial pressure of nitrogen, [1 atm = 101325 Pa]
`P_(N_2) = 0.8atm = 81060 Pa` ltbagt Using Henry's law equation
`chi=P/K_(H)`
`:.` Amount fraction of `O_(2)` in water
`chi_(O2) = "in" P_(O2)/K_(H(O2)) = (20265)/(2 xx 10^(9)) = 10.13 xx 10^(-6) `
Amount fraction of `N_(2)` in water
`chi_(N2) = "in" P_(N2)/K_(H(N2)) = (81060)/(5 xx 10^(9)) = 16.212 xx 10^(-6) `
`:. ("Amount of dissolved oxygen")/("Amount of dissolved nitrogen")= (10.13 xx 10^(-6))/(16.212 xx 10^(-6)) =(0.62)`
Hence, amount percent `O_(2)`=(0.62/1.62) xx (100) =(38.27%)`
Amount percent of `N_(2)=100-38.27 = 61.73%`