What is the mole ratio of benzene (C6H6) to n-octane in the vapor above a solution of 15.0% benzene and 85.0% n-octane by mass at 25 degrees Celcius? the vapor pressures of n-octane and benzene are 11 torr and 95 torr.

Answer 1

15% benzene means 15g benzene/(15g benzene + 85 g octane).
15/molar mass benzene = mols benzene.
85/molar mass octane = mols octane.

Calculate X_benzene
Calculate X_octane.
Then pBen = XBen*P^o_benzene
and pOct = XOct*P^o_octane

Then Ptotal = Pbenzene + Poctane

Answer 2

here is what I did (check my work, because something isnt right):

15g C6H6/78.12gC6H6 = .19 mol C6H6
85g C8H18/114.26g C8H18= .74 mol C8H18

X-benzene: .19/.93= .204
X-octane: .74/.93= .796

pBen: .204*95=19.38
pOctane: .796*11=8.756

19.38+ 8.756= 28.136

Answer 3

That looks ok to me. You haven't been consistent with significant figures. If your database and prof are picky about that you need to go through and make some changes.

Answer 4

The database said the answer was 2.23 ??

Answer 5

That' because you and I didn't finish the question. Go back and look at the problem. It asks for the benzene/octane RATIO. So take pbenzene/poctane. Using your numbers that is 19.38/8.756 = 2.21 and I suspect the difference (2.21 vs 2.23) is one of significant figures through the computation.

Answer 6

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