If #"45.0 g"# of methanol, #"22.0 g"# of acetone, and #"120.0 g"# of water are combined, what is the mol fraction of methanol?

I got #chi_("MeOH") = 0.166#.

Well, it's like it says. A mole fraction is a fraction in terms of mols. A fraction by definition is the contribution one thing makes to the whole.

Thus, the fraction of particles in something is

#"Number of particle i"/"Number of particles 1, 2, 3, 4, 5, . . . "#

In terms of mols #n_k# of substance #k#, the mole fraction #chi_k# is

#chi_k = n_k/(n_1 + n_2 + . . . )#

So first convert everything to mols...

#n_("MeOH") = 45.0 cancel("g CH"_3"OH") xx ("1 mol CH"_3"OH")/(32.0416 cancel"g") = "1.404 mols methanol"#

#n_("ac") = 22.0 cancel("g C"_3"H"_6"O") xx ("1 mol C"_3"H"_6"O")/(58.0794 cancel"g") = "0.3788 mols acetone"#

#n_(w) = 120.0 cancel("g H"_2"O") xx ("1 mol H"_2"O")/(18.015 cancel"g") = "6.661 mols water"#

Then the mole fraction of methanol is...

#color(blue)(chi_("MeOH")) = n_("MeOH")/(n_"MeOH" + n_"ac" + n_w)#

#= "1.404 mols"/("1.404 mols CH"_3"OH" + "0.3788 mols C"_3"H"_6"O" + "6.661 mols H"_2"O")#

#= color(blue)(0.166)#

to three significant figures.

CHALLENGE: If the mole fraction of acetone is known to be #0.0449#, what is the mole fraction of water? You should be able to do this in less than one minute if you know the trick.

If you figure that out, highlight below:

#color(white)(chi_("H"_2"O") = 0.789)#