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.

Calculate the proporation of O _(2) N_(2) dissolved in water at 298K.

1.	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.
Answer» Solution :`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`