Question

He henry's law constant of oxygen in water at 25 °c is 773 atm mol-1kg h2o. calculate the molality of oxygen in water under a partial pressure of 0.20 at

Answer 1

Henry law:

P = KC
P = solvent vapour pressure
K = Henry constant 
C = conc of solute

So,   0.2 = 773 C
so, C =  0.2 / 773 =   2.5 x 10^{-4} molal