O2is bubbled through water at 293K. Assuming that O2 exerts a partial pressure of 0.98 bar, calculate the solubility of O2 in grams per litre. The value of Henry's Law constant (KH) for O2 is 34.84 kbar
pO2=KH*χO2
Thus
χO2=pO2/KH =0.098/(34.84*103) =2.81*10-5
No. of moles of H2O in 1L =55.55
χO2 = nO2/(nO2 + nH2O) = nO2/(nO2 +55.55)
APPROXIMATELY
χO2 =nO2/55.55
nO2 =2.81*10-5*55.55 =1.56*10-3 mol
Thus solubility of O2 is = 1.56*10-3 *32 g/L = 0.05 g/L
Thus
χO2=pO2/KH =0.098/(34.84*103) =2.81*10-5
No. of moles of H2O in 1L =55.55
χO2 = nO2/(nO2 + nH2O) = nO2/(nO2 +55.55)
APPROXIMATELY
χO2 =nO2/55.55
nO2 =2.81*10-5*55.55 =1.56*10-3 mol
Thus solubility of O2 is = 1.56*10-3 *32 g/L = 0.05 g/L