Chapter 4:
Chemical
Composition
Part 5
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4-1
Percent by Mass of Solute

Solution concentration is often expressed as the
mass percent of solute:
mass of solute
Percent Mass Solute =
100
total mass of solution
mass of solute + mass of solvent
4-2
Activity: Percent by Mass of Solute

What is the mass percent of NaCl in a solution that
is prepared by adding 10.0 g NaCl to 50.0 g water?
mass of solute
Percent Mass Solute =
100
total mass of solution
10.0 g NaCl
Percent Mass NaCl =
100 = 16.7 %
60.0 g solution
4-3
Molarity (M)

Another common way to express the
concentration of a solution is in molarity units:
A
B
C
B  AC
B
C
A
moles solute
Molarity =
liters of solution
Moles of solute  molarity  (liters of solution)
moles of solute
liters of solution 
molarity
4-4
Activity: Preparing a CuSO4 Solution



6.25 grams (0.0250 mol) of CuSO45H2O is added
to a 250-mL volumetric flask.
Water is added to the mark so that the total
volume is 250.0 mL.
What is the molarity of this solution?
0.0250 mol
Molarity =
= 0.100 M
0.250 L
Figure 4.19
4-5
Activity: Molarity

How many moles of NaCl are in 1.85 L of a
0.25 M NaCl solution?
moles NaCl = 1.85 L × 0.25 mol/L = 0.46 mol
moles solute
Molarity =
liters of solution
Moles of solute  molarity  (liters of solution)
moles of solute
liters of solution 
molarity
4-6
Activity: Solution Concentration

Bluestone is copper(II) sulfate pentahydrate,
CuSO4•5H2O, with a molar mass of 249.7 g/mol. A
sample of pond water was found to have a
concentration of 6.2  10-5 M copper(II) sulfate. If
the pond has a volume of 1.8  107 L, then what mass
of bluestone did the farmer add to the pond as an
algicide?
4-7
Activity Solution: Solution Concentration

Bluestone is copper(II) sulfate pentahydrate, CuSO4•5H2O, with
a molar mass of 249.7 g/mol. A sample of pond water was found
to have a concentration of 6.2  10-5 M copper(II) sulfate. If the
pond has a volume of 1.8 x 107 L, then what mass of bluestone
did the farmer add to the pond as an algicide?
-
5
6.2  10 mol
249.7 g
1.8  10 L 

= 2.8  1015 g CuSO4 5H 2O
1L
1 mol
5
7
4-8
Dilution

Suppose you want to dilute a 0.25 M solution to a
concentration of 0.025 M. What are some ways to
do this?
4-9
Dilution
Figure 4.21
4-10
Describe this process
Figure from p. 163
4-11
Dilution

Solute molesinitial= Solute molesfinal



Moles = Molarity × Volume
Moles = mol/L × L
MinitialVinitial = MfinalVfinal
4-12
Activity: Dilution

MinitialVinitial = MfinalVfinal

What is the concentration of a solution prepared
by adding water to 25.0 mL of 6.00 M NaOH to a
total volume of 500.0 mL?
4-13
Activity: Dilution

MinitialVinitial = MfinalVfinal

What is the concentration of a solution prepared
by adding water to 25.0 mL of 6.00 M NaOH to a
total volume of 500.0 mL?
MinitialVinitial = MfinalVfinal
6.00M  0.0250L  M final0.500L
6.00 M  0.0250 L
M final =
= 0.300 M
0.500 L
4-14
Activity: Dilution

If 42.8 mL of 3.02 M H2SO4 solution is diluted to a
final volume 500.0 mL, what is the molarity of the
diluted solution of H2SO4?
4-15
Activity Solution: Dilution

If 42.8 mL of 3.02 M H2SO4 solution is diluted to a
final volume 500.0 mL, what is the molarity of the
diluted solution of H2SO4?
MiniVini = MfinVfin
3.02 M × 42.8 mL = Mfin× 500.0 mL
3.02 M  42.8 mL
M dilfin =
= 0.259 M
500.0 mL
4-16
Activity: Dilution

What is the concentration of a solution prepared by
diluting 35.0 mL of 0.150 M KBr to 250.0 mL?
4-17
Activity Solution: Dilution

What is the concentration of a solution prepared by
diluting 35.0 mL of 0.150 M KBr to 250.0 mL?
MiniVini = MfinVfin
0.150 M × 35.0 mL = Mfin × 250.0 mL
0.150 M  35.0 mL
M dil =
= 0.0210 M
fin
250.0 mL
4-18
Homework
Page 163Problem
4.99
4.101a,d
4.103
4.106a
4.108a 4.114a
4-19