Question

The density of a 15.00% by mass aqueous solution of acetic acid, CH3COOH, is 1.0187g/mL. What is (a) the molarity? (b) the molality? (c) the mole fraction of each component?

Answer 1

density = 1.0187 g/mL.
mass 1000 mL = 1.0187 x 1000 mL = ??
How much of that is acetic acid? 15%; therefore, ?? x 0.15 = mass acetic acid = xx.
How many mols is that?
xx g acetic acid/molar mass acetic acid = yy
Molarity = yymols/L = yy M.

molality = mols solute/kg solvent
mols you have from above. kg solvent is mass of 1 L solution which is 1018.7 g- mass acetic acid = mass H2O remaining.

mol fraction.
mols acetic acid = from above.
mols H2O = g H2O/molar mass H2O.

mol fraction acetic acid = mols acetic acid/sum mols acetic acid + mols H2O.

mols fraction H2O = etc.

Check my work. Check my thinking.

Answer 2

thanks so much for your help!!
i think i understood everything you said...but just in case am i on the right track with these answers?

molarity = 2.545M
molality = 2.939m
mole fraction: acetic acid = .05030
H20 = .9497

Answer 3

I remember working the molarity and molality and those are correct if my memory isn't failing me. I just worked the mole fraction one, but rounded on molar masses etc, and your numbers agree at least to three significant figures. So I think you have it ok. Glad to help. Come back anytime.

Answer 4

a 7.50 by mass aqueous ammonia, NH3,solution has a density of 0.973 g/mol. Calculate molality and molarity? Need your ASAP please :(