# chemistry

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.

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1. 15% benzene means 15g benzene/(15g benzene + 85 g octane).
15/molar mass benzene = mols benzene.
85/molar mass octane = mols octane.

Calculate Xbenzene
Calculate Xoctane.
Then pBen = XBen*Pobenzene
and pOct = XOct*Pooctane

Then Ptotal = Pbenzene + Poctane

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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

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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.

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4. The database said the answer was 2.23 ??

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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.

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