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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