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October 23, 2014

October 23, 2014

Posted by **keely** on Saturday, November 10, 2007 at 9:14am.

- chemistry repost!! -
**drwls**, Saturday, November 10, 2007 at 11:54amMolarity 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

- chemistry repost!! -
**DrBob222**, Saturday, November 10, 2007 at 3:27pmTo 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.

- chemistry repost!! -
**keely**, Saturday, November 10, 2007 at 3:56pmwow thanks! its so very clear!

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