We're asked to find the **molarity** and **mole fraction of acetic acid** in a solution.

Well, the molarity is given as #2M#...so I will also find the **molality** of the solution.

We know the density of the #"CH"_3"COOH"# solution is #1.2# #"g/mL"#, which is the same as #color(green)(1200# #color(green)("g/L"#.

Let's assume we have #1# #"L"# of solution..

Then, there are #2# #"mol CH"_3"COOH"# in the solution, because the given molarity value says there are two moles of solute per liter of solution.

Using the molar mass of acetic acid, we can find the number of **grams** of acetic acid present (which we'll use to find the quantity of water (solvent) present):

#2cancel("mol CH"_3"COOH")((60.052color(white)(l)"g CH"_3"COOH")/(1cancel("mol CH"_3"COOH"))) = color(red)(120.104# #color(red)("g CH"_3"COOH"#

Now, using the given density of the solution, we can find the number of *grams* of solution; we can then subtract the grams of solute from that to find the mass of water:

Since we assumed #1# #"L soln"#, according to the density value there are #color(green)(1200# #color(green)("g soln"#, so the mass of water present is

#color(green)(1200color(white)(l)"g soln") - color(red)(120.104color(white)(l)"g CH"_3"COOH") = color(purple)(1080# #color(purple)("g H"_2"O"#

or

#color(purple)(1.080color(white)(l)"kg H"_2"O"#

Thus, the **molality** of the solution is

#"molality" = "mol solute"/"kg solvent" = (2color(white)(l)"mol CH"_3"COOH")/(color(purple)(1.080color(white)(l)"kg H"_2"O")) = ul(1.85m#

Lastly, let's find the **mole fraction of acetic acid** of the solution.

What we can do is calculate the number of moles of **water** present using its molar mass (#18.015# #"g/mol"#) and mass present (#color(purple)(1080# #color(purple)("g"#):

#color(purple)(1080)cancel(color(purple)("g H"_2"O"))((1color(white)(l)"mol H"_2"O")/(18.015cancel("g H"_2"O"))) = color(orange)(59.9# #color(orange)("mol H"_2"O"#

The mole fraction of acetic acid is thus

#chi_("CH"_3"COOH") = ("mol CH"_3"COOH")/("mol CH"_3"COOH" + "mol H"_2"O")#

#= (2color(white)(l)"mol CH"_3"COOH")/(2color(white)(l)"mol CH"_3"COOH" + color(orange)(59.9color(white)(l)"mol H"_2"O")) = color(blue)(0.0323#