posted by Jude .
What is the mole fraction of benzene C6H6 in the VAPOR PHASE from a solution consisting of a mixture of benzene and n-hexame ch3(ch2)4ch3 at 27celsius, if the total pressure above the solution is 135.2 torr? The vapor pressures of benzene and n-hexame at 27 celsius are 103 and 161.7 torr respectively. Hint: Assuming these compounds form nearly ideal solutions when mixed together, determine the mole fraction of benzene in the liquid phase first, note that the mole fraction of benzene in the vapor phase will be its vapor pressure divided by total pressure.
I get x for benzene to be 0.451 but its incorrect. What should i be doing differently? Thanks
26.5/58.7 = 0.451 = Xbenzene
32.2/58.7 = 0.548 = Xhexane
We can check that out.
0.451*103 = 46.45 mm benzene
0.548*161.7 = 88.61 mm hexane
total = 135.1 which is close.
Xbenzene in vapor phase = Pbenzene/total P = 46.45/135.2 = ??
See the correction I made to the methanol propanol post of Saturday. Look for your post at about 4:50 pm or so.