Asked by Lori
The mole fraction of an aqueous solution of potassium perchlorate is 0.125. Assuming that the density of this solution is 1.20g mL-1, calculate the following:
(A) the mass percent of potassium perchlorate in the solution.
(B) the molarity of potassium perchlorate in the solution.
(A) the mass percent of potassium perchlorate in the solution.
(B) the molarity of potassium perchlorate in the solution.
Answers
Answered by
DrBob222
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.
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.
Answered by
Lori
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
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
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