posted by Jude .
What is the mole fraction of methanol CH3OH, in the vapor phase from a solution consisting of a mixture of methanol and propanol CH3CH2CH2OH at 40celsius, if the total pressure above the solution is 131 torr? The vapor pressures of methanol and propanolare 303 and 44.60 torr respectively. Assuming these compounds form nearly ideal solutions when mixed together.
Propanol = pr
Methanol = me
Pure pr = 44.6
Pure me = 303
The difference between pure pr and pure me is 303-44.6 = 258.4
delta P from pure pr = 131-44.6 = 86.4
delta P from pure me = 303-131 = 172.0
86.4/258.4 = mole fraction me.
172.0/258.4 = mole fraction pr.
The best way to see this is to draw it out on a piece of paper.
To check the answer I did this.
0.3344 x 303 = 101.32 mm
0.6656 x 44.6 = 29.68 mm
Total 131.000. Of course, that many significant figures are not allowed. You would round to 0.334 and 0.666 which gives 130.9 which rounds to 131. I hope this helps.
I followed this procedure but i was told the answer was incorrect. Any help would be appreciated.
The above procedure is correct as far as I went BUT the numbers calculate the mole fraction of me and pr in the liquid phase. To find the mole fraction of each in the vapor phase,
Xme vapor phase = Pme/total P
Xpr vapor phase = Ppr/total P.