Posted by Lori on .
Posted by Lori on Tuesday, December 14, 2010 at 5:38pm.
The mole fraction of an aqueous solution of potassium perchlorate is 0.125. Assuming that the density of this solution is 1.20g mL1, calculate the following:
(A) the mass percent of potassium perchlorate in the solution.
(B) the molarity of potassium perchlorate in the solution.
chem  DrBob222, Tuesday, December 14, 2010 at 7:52pm
mole fraction KClO4 = 0.125 and mole fraction water = 0.875. If we take enough sample to have 1.00 total mol soln, we will have 0.125 moles KClO4 or 0.125 x molar mass KClO4 in grams. 0.875 moles H2O = 0.875 x molar mass water = grams water.
grams soln = grams water + grams KClO4.
mass % w/w = (mass KClO4 in grams/mass soln)*100
b)Convert g KClO4 (in previous problem) to moles. moles = grams/molar mass.
Use the density to convert g soln to L. Then M = moles/L of soln.
Post your work if you get stuck.
chem  Lori, Tuesday, December 14, 2010 at 10:52pm
Ok I got (A) to work out for me 52.4%
but i'm stuck on (B)
moles= grams/molar mass
1.20g/138.547= 8.66x10^3 ( this cant be right)
then M = moles/L 8.66x10^3/1.2x10^3= 7.22
I got 7.22 but the right answer is 4.53M

chemistry needs another look plz 
DrBob222,
but i'm stuck on (B)
moles= grams/molar mass
1.20g/138.547= 8.66x10^3 ( this cant be right)
You are using density/molar mass to find moles which isn't correct. From part one, you already know moles = 0.125 moles KClO4 but you can calculate it again if you wish from 17.318 g KClO4/138.5486 = 0.125.
The volume of the solution from part 1 was 33.081 grams. The density is 1.20 g/mL; therefore, volume = mass/density = 33.081/1.20 = 27.5675 mL or 0.0275675 L. So M = moles/L = 0.125 moles/0.0275675L = 4.5343M which rounds to 4.53M to three s.f.
then M = moles/L 8.66x10^3/1.2x10^3= 7.22
I got 7.22 but the right answer is 4.53M