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?

Answer 1

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.

Answer 2

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Answer 3

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?