At what partial pressure, oxgyenwill 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 .

At what partial pressure, oxgyenwill have a solubility of "0.05 g L"^(

1.	At what partial pressure, oxgyenwill 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 .
Answer» <P> SOLUTION :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."`