Calculate the proporation of `O _(2) N_(2)` dissolved in water at 298K. When air containing `20% O _(2)and 80% N_(2)` by volume is in equilibrium with water 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

`C_(1) V_(1) = C_(2) V_(2)`
`6M (V_(1)) = 0.25 M xx 500 ml`
`V _(1) = (0.25 xx 500)/(6)`
` V _(1) = 20.83mL`