Question

The mole fraction of acetic acid (CH3COOH) in an aqueous solution is 0.675. Calculate the molarity of the acetic acid, if the density of the solution is 1.0266 g mL-1.

Answer 1

Molarity is moles of solute (acetic acid) per liter of solution. I will call acetic acid X to save typing.

[Moles X/liter solution] = [Moles X/Moles H2O]x [Moles H2O/liter]
=[Moles X/Moles H2O]x[MolesH2O/g H2O][gH2O/g solution][g solution/liter]
=(.325/.675)(1/18)MolesX/g/H2O[gH2O/g solution][1026.6 g/l]
You still need the mass fraction of H2O in the solution. That can be computed from
the mole fractions and molecular masses and turns out to be 0.1644 gH2O/gsolution

Answer 2

To give a different take on this, here is another way of approaching the problem. I don't claim it is better or easier. Let X = mols acetic acid; Y = mols H2O. (Two equations and two unknowns).
mol fraction acetic acid = X/(X+Y) = 0.675
solve for X in terms of Y. I have
X = 2.077*Y

(second equation)
mol fraction water = 1-X = 0.325 = Y/(X+Y)
solve for Y in terms of X. I have
0.481*X
Plug 0.481X in for Y and I get 1 mol for X; plug this value into the second equation to obtain Y = 0.481 mol.

Then 1 mol X = 1 mol CH3COOH = 60 g (You should check this as I estimated.)
0.481 mol Y = 0.481 mol H2O = 8.658 g
Total solution = 68.658 g.
Use density to convert this to volume.
Volume = mass/d = 66.879 mL.
M = mols/L = 1 mol/0.066879 = 14.95 M which probably should be rounded to 15 M since it appears we have only 3 significant figures.
Check my thinking. Check my work.

Answer 3

wow thanks! its so very clear!