Question

1.The solubility of O₂ in water is 0.590 g/L at an oxygen pressure of around 15.5 atm. What is the Henry's law constant for O₂ (in units of mol/L

Answer 1

given
Po2= 15.5 atm
solubility of O2 = 0.590 g/l
molar mass of O2 = 32 g/mol

solubility in mol/l = solubility in g/l/molar mass

So solubility of O2 =0.590 g/l/32 g/mol
= 0.01844 mol/L

Henry's law formula

solubility S= K* Po2

Henry's constant K = S/Po2 = 0.01844 mol/L/15.5 atm
= 1.19 x10^-3 mol/l-atm

Answer : 1.19 x10^-3 mol/l-atm