AP CHEM NOTES
CONCENTRATIONS SAMPLE PROBLEMS
CONCENTRATION EQUATIONS:
Molarity:
% solute:
Molality:
parts per million:
Normality:
N = nM
parts per billion:
+
n=
# H in acids
−
# OH in bases
−
# e in redox
mole fraction:
SAMPLE PROBLEMS:
1) Given a 1.75 M NaOH solution having a density of 1.040 g/mL, determine the molality, % NaOH, normality and
mole fraction of NaOH.
Make an assumption for a starting point.
By the assumption, and that there is 1.75 M,
figure the moles of solute.
Figure the mass of solute by multiplying by its
molar mass.
Determine the volume of the solution—this is
from your assumption.
Assume 1 L of solution.
1.75 mol NaOH
70.0 g NaOH
1000 mL soln
Using density, find the mass of the solution.
m = (1.040 g/mL)(1000 mL) = 1040. mL solution
Find the mass of water by subtracting the mass
of the solution by the mass of NaOH.
Find the moles of H2O by dividing its mass by
its molar mass.
SUMMARY OF DATA:
1.75 mol NaOH = 70.0 g NaOH
53.8 mol H2O = 970. g H2O
mass H2O = 1040. g soln – 70.0 g NaOH = 970. g H2O
53.8 mol H2O
mass soln = 1040. g soln
vol. soln = 1000 mL
Finding molality:
Finding % NaOH:
Finding normality:
Finding mole fraction:
(
)
(
N = nM = (1)(1.75 M) = 1.75 N NaOH
)
2) Concentrated sulfuric acid, H2SO4, has a molality of 189.36 m and a density of 1.84 g/mL. Find the molarity,
normality, % H2SO4 and mole fraction H2SO4.
Make an assumption for a starting point.
By the assumption, and that the solution is
189.4 m, figure the moles of solute.
Figure the mass of solute by multiplying by its
molar mass.
From the assumption, determine the mass of
water.
From the mass, find the moles of water.
Find the mass of solution by adding the mass
of solute and solvent.
Assume 1 kg of water.
189.36 mol H2SO4
18570 g H2SO4
1000 g H2O
55.49 mol H2O
mass soln = 18570 g solute + 1000 g solvent = 19570 g soln
Using density, find the volume of the solution.
SUMMARY OF DATA:
189.36 mol H2SO4 = 18570 g H2SO4
55.49 mol H2O = 1000 g H2O
mass soln = 19570 g soln
vol. soln = 10600 mL soln
Finding molarity:
Finding normality:
Finding % H2SO4:
N = nM = (2)(17.9 M) = 35.8 N H2SO4
(
)
(
Finding mole fraction:
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)
3) Given a 18.40% Mg(NO3)2 solution with a density of 1.045 g/mL, find the molarity, molality, and mole fraction of
Mg(NO3)2.
Make an assumption for a starting point.
By the assumption, and that the solution is
18.40%, figure the mass of solute and the mass
of solvent.
Assume 100 g solution.
18.40 g Mg(NO3)2
81.60 g H2O
Determine the moles of the solute and solvent.
0.1240 mol Mg(NO3)2
4.528 mol H2O
From the assumption, you know the mass of
solution.
mass soln = 100 g
Using density, find the volume of the solution.
SUMMARY OF DATA:
0.1240 mol Mg(NO3)2 = 18.40 g Mg(NO3)2
4.528 mol H2O
= 81.60 g H2O
mass soln = 100 g soln
vol. soln = 95.69 mL soln
Finding molarity:
Finding molality:
Finding mole fraction:
4) Given a NaOH solution with XNaOH = 0.1000, find the molality and %NaOH.
Since we are not finding molarity nor normality, we don’t need to find the volume of the solution.
Make an assumption for a starting point.
By the assumption, determine the moles of the
solute and solvent.
Determine the mass of the solute and solvent.
Determine the mass of the solution.
SUMMARY OF DATA:
0.1000 mol NaOH = 4.000 g NaOH
0.9000 mol H2O = 16.22 g H2O
Assume 1 mole solution.
0.1000 mol NaOH
0.9000 mol H2O
4.000 g NaOH
16.22 g H2O
mass soln = 4.000 g + 16.22 g = 20.22 g soln
mass soln = 20.22 g
Finding molality:
Finding % NaOH:
(
)
(
)