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Calculate the molarity and mole fraction of acetone in a 1.29 m solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

  • chemistry -

    1.29 m means 1.29 moles/kg solvent.
    molar mass acetone about 58
    molar mass ethanol about 46
    You need to redo all of this since I've estimated some of these numbers.
    Take enough solution to have 1.29 moles which means you take 1 kg solution.
    1.29 x 58 = about 75 grams acetone.
    The solution has a mass of 1000 kg ethanol + 75 g acetone = 1075 total. What volume is that?
    The solution has a density of about 0.788 (since the density of acetone and ethanol are almost the same and we are assuming the volumes are additive, then the density of the solution is very close to 0.788.
    volume = mass/density = 1075/0.788 = ??
    Change that to L, then M = mols/L.

    To convert to mole fraction.
    moles acetone = 1.29
    moles ethanol = 1000 g/molar mass
    mole fraction acetone = moles acetone/total moles.

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