chemistry

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a mixture is prepared using equal masses of two volatile liquid benzene, C6H6 and ethanol C2H5OH. what is the mole fraction of benzene in this mixture?

  • chemistry -

    I would pick a number for the masses.
    The convert masses to mols.
    mols fraction X = molsX/total # of mols.
    Post your work if you get stuck.

  • chemistry -

    DrBob222 you are the best thanks for all you help...you great. thanks

  • chemistry -

    ok, so I assumed the masses to equal 5.00 g

    molar mass of benzene is =78.12 g/mol

    molar mass of ethanol is =46.08g/mol

    moles of C6H6=
    5.00g/78.12g/mol=0.064004moles

    moles of C2H5OH=
    5.00 g/48.08g/mol=0.10851g/mol

    so mole fraction of benzene should equal to=
    0.064004mol/(0.064004+0.10851)=0.371mol

    so the mole fraction of benzene is 0.371 mol

    Is my approach to the problem correct?

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