A student was performing an extraction procedure of o-Xylene from water into octanol. The total mass of o-Xylene was 0.10 gram (molecular weight 106.2 g/mole). The volume of water was 990mL and volume of octanol 10 mL. There is a head space of 200 mL above the solution and it is sealed. Temperature was 25°C. The Henry’s Law constant for o-Xylene is 0.2Matm-1 and logKOW is 3.12. Calculate the concentrations of o-Xylene in the air above the solution in ?g/m3, concentration in water and octanol in mg/L.

^{-1}) * 0.1 (g) / 106.2 (g/mole) * 1/1 (L) , since total volume of liquid is 1L.

^{-1}) * 0.2 (L) * 1/298 (K) / 0.082 (L*atm mol

^{-1}K

^{-1})

^{-4}g (in 200cc). Its concentration in g m

^{-3}is

^{-4}(g) * 1/200 (cm

^{3}) * 10

^{6}(cm

^{3}/ m

^{3}), or 0.8185 g m

^{-3}.

_{OW}= 3.12. Volume of water and xylene are 990 mL and 10mL respectively.

^{-1}.

_{OW}= 3.12 , hence Conc in octanol / Conc in water is 10

^{3.12}= 1318.257.

^{-1}, then

^{-1}, and the concentration of xylene in octanol is 100.9335 mg L

^{-1}. These values were obtained by multiplying (1-x) and x respectively with 101.01. The units of concentration did not make a difference, as we were using ratios of concentration in the partition coefficient equation.

## Comments

remove that '?' and you get grams per cubic meter, which would be the concentration of xylene in the air.