Dry air contains 79% `N_2` and 21% `O_2`. Determine the proportion of `N_2 and O_2`, dissolved in water at 1 atm pressure. Henry's law constant for `N_2 and O_2 " in " H_2O " are " 8.54 xx 10^4` atm and `4.56 xx 10^4` atm respectively.

Explanation

Total pressure of air over water = 1 atm
Partial pressure of `N_2, P_(N_2) = (1xx79)/1000 = 0.79` atm
Partial pressure of `O_2`
`P_(O_2) =(1xx21)/100 = 0.21` atm
Applying Henry.s law, `p=K_H *x`
`x_(N_2) = (p_(N_2))/(K_(N_2)) = (0.79 " atm ")/(8.54xx10^4 "atm" ) = 9.25xx10^(-6)`
`x_(O_2)=(p_(O_2))/(K_(O_2)) = (0.21 " atm ")/(4.56xx10^4 " atm " ) = 4.60xx10^(-6)`
Proportion of `N_2 and O_2 = 9.25xx10^(-6) : 4.6xx10^(-6)=2`