At what partial pressure, oxgyen will have a solubility of `"0.05 g L"^(-1)` in water at 293 K ? Henry's constant `(K_(H))` for `O_(2)` in water at 293 K is 34.86 kbar. Assume the density of the solution to be same as that of the solvent .

Explanation

Calculation of mole fraction `(x_(O_(2)))`
Mass of 1 L of solutio = 1000 g`" "(because d ="1 g mL"^(-1))`
`therefore" Mass of solvent (water) "=1000g-0.05 g~=1000g`
`therefore" "n_(H_(2)O)=("1000 g")/("18 g mol"^(-1))="55.5 moles, "n_(O_(2))=("0.05 g")/("32 g mol"^(-1))=1.56 xx10^(-3)" mole"`
`therefore" "x_(O_(2))=(n_(O_(2)))/(n_(O_(2))+n_(H_(2)O))~=(n_(O_(2)))/(n_(H_(2)O))=(1.56xx10^(-3))/(55.5)=2.81xx10^(-5)`
Calculation of partial pressure. Applying Henry's law,
`p_(O_(2))=K_(H)xx x_(O_(2))=(34.86xx10^(3)" bar")xx(2.81xx10^(-5))="0.98 bar."`