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.

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

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.

wow thanks! its so very clear!

To calculate the molarity of acetic acid, we need to use the mole fraction and the density of the solution. Here are the steps to follow:

1. Convert the mass density of the solution from grams per milliliter (g/mL) to grams per liter (g/L). Since there are 1000 mL in 1 L, we can multiply the given density by 1000 to get the density in g/L.

1.0266 g/mL x 1000 = 1026.6 g/L

2. Determine the molar mass of acetic acid (CH3COOH) by summing up the atomic masses of each element.

The atomic masses of carbon (C), hydrogen (H), and oxygen (O) are approximately:

C: 12.01 g/mol
H: 1.01 g/mol
O: 16.00 g/mol

Calculating the molar mass of acetic acid:

(12.01 x 2) + (1.01 x 4) + 16.00 + 1.01 = 60.05 g/mol

3. Use the mole fraction to calculate the moles of acetic acid present in the solution. The mole fraction (X) is the ratio of the moles of acetic acid (n(CH3COOH)) to the total moles of the solution (n(total)).

X = n(CH3COOH) / n(total)

Rearranging the equation to solve for n(CH3COOH):

n(CH3COOH) = X x n(total)

Since the mole fraction of acetic acid is given as 0.675, we can substitute this value into the equation.

n(CH3COOH) = 0.675 x n(total)

4. Convert the moles of acetic acid to molarity. Molarity (M) is defined as the number of moles of solute divided by the volume of the solution in liters.

M = n(CH3COOH) / V

Since we don't have the volume of the solution, we will leave the molarity in terms of moles per liter (mol/L).

Therefore, the molarity of the acetic acid is 0.675 mol/L