Mole fractions in a liquid mixture:

Acetone, C3H6O, and ethyl acetate, C4H8O2, are organic liquids often used as solvents. At 30 degrees C, the vapor pressure of acetone is 285 mm Hg and the vapor pressure of ethyl acetate is 118mm Hg. the solution is prepared at 30 degrees C by dissolving 27.0 g of acetone in 22.5 g of ethyl acetate.

A. What is the mole fraction of each component in the liquid mixture?

B. What is the mole fraction of each comment in the vapor at 30 degrees C?

How do I even START this problem? I am so confused.

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  1. Calculate moles acetone.
    Calculate moles ethyl acetate.
    Add mole acetone to moles ethyl acetate for total moles.
    Now you are set up.
    A. liquid
    mole fraction acetone = moles acetone/total moles.
    mole fraction ethyl acetate = moles ethyl acetate/total moles.

    Part B. vapor
    vapor (partial) pressure acetone = mole fraction acetone x normal vapor pressure acetone.

    vapor (partial) pressure ethyl acetate = mole fraction ethyl acetate x normal vp ethyl acetate.

    add partial pressure acetone in vapor to partial pressure ethyl acetate in vapor to obtain total pressure.
    Then
    mole fraction acetone = partial pressure acetone/total pressure
    mole fraction ethyl acetate = partial pressure ethyl acetate/total pressure.

    Check my thinking.

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  2. When I did that I got this:

    2.6 for acetone mole fraction
    1.4 for ethyl acetate mole fraction

    119 acetate vapor
    30 ethyl acetate vapor

    None of those answers were correct. :[

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