Using Henry's Law, we get the partial pressure of the gas above the solution as
p = 0.2 (atm M-1) * 0.1 (g) / 106.2 (g/mole) * 1/1 (L) , since total volume of liquid is 1L.
This value of p is transferred to the gas equation PV = (g/M)RT to determine the weight g of xylene in the headspace gas. The number of moles "n" has been replaced by weight (g) divided by molecular weight (M).
I did not calculate the value of p in the above equation, because p * M = 0.02 and simplifies things as we go along.
g now becomes pVM/RT, or
g = 0.02 (atm*g*mol-1) * 0.2 (L) * 1/298 (K) / 0.082 (L*atm mol-1 K-1)
After calculations, g = 1.63693 * 10-4 g (in 200cc). Its concentration in g m-3 is
1.63693 * 10-4 (g) * 1/200 (cm3) * 106 (cm3 / m3), or 0.8185 g m-3.
Since g is ~ 1000 times less than 0.1g, we ignore this in the partition calculations. Total mass of xylene is 0.1 g (100mg) and log KOW = 3.12. Volume of water and xylene are 990 mL and 10mL respectively.
If x is the concentration of xylene in octanol, then (1-x) is the concentration in water, since extraction is from water to octanol, and assuming that 1M was the original concentration in water.
Original concentration of xylene in water = 100 (mg) /0.99 (L) = 101.01 mg L-1.
log KOW = 3.12 , hence Conc in octanol / Conc in water is 103.12 = 1318.257.
1318.257 = x / (1-x). Solving for x, we get
x = 1318.257 / 1319.257 = 0.999242, and
1-x = 0.000758
If the original concentration of xylene in water was 101.01 mg L-1, then
The new concentration of xylene in water is 0.07656 mg L-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.