Calculate the proportion of `O_(2)` and `N_(2)` dissolved in water at 298 K. When air containing `20% O_(2) " and " N_(2)` by volume is in equilibrium with it at 1 atm pressure. Henry's law constants for two gases are `K_(H)(O)_(2) = 4.6 xx 10^(4) atm " and " K_(H)(N_2) = 8.5 xx 10^(4)` atm.

Explanation

Total pressure = 1 atm
`P_(N_2)= ((80)/(100))xx` total pressure `= (80)/(100)xx 1 atm = 0.8` atm
`P_(O_2)= ((20)/(100))xx1= 0.2 atm`
According to Heney.s Law
`P_("solute")= K_(H)X_("solute in solution ")`
`therefore P_(N_2)= (K_H)_("Nitrogen")xx` mole fraction of Nitrogen in solution
`(0.8)/(8.5xx10^(4))= X_(N_2)`
`X_(N_2)= 9.4xx10^(-6)`
Similarly, `X_(O_2)= (0.2)/(4.6xx10^(4))`
`= 4.3xx 10^(-6)`.