Chapter 8B
Solution
Concentrations
1
CHAPTER OUTLINE

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Concentration Units
Mass Percent
Using Percent Concentration
Molarity
Using Molarity
Dilution
Osmolarity
Tonicity of Solutions
2
CONCENTRATION
UNITS
 The amount of solute dissolved in a certain amount
of solution is called concentration.
Concentration =
amount of solute
amount of solution
 Three types of concentration units will be studied in
this class:
Mass Percent: (m/m) and (m/v)
Molarity
3
MASS PERCENT
 Mass percent (% m/m) is defined as the mass of
solute divided by the mass of solution.
mass of solute
Mass % (m/m) =
x100
mass of solution
mass of solute +
mass of solvent
4
MASS/VOLUME
PERCENT
 Mass/Volume percent (% m/v) is defined as the mass
of solute divided by the volume of solution.
mass of solute
Mass % (m/v) =
x100
volume of solution
5
Example 1:
What is the mass % (m/m) of a NaOH solution that
is made by dissolving 30.0 g of NaOH in 120.0 g of
water?
Mass of solution = 30.0 g + 120.0 g = 150.0 g
30.0 g
Mass % (m/m)=
x100 = 20.0 %
150.0 g
6
Example 2:
What is the mass % (m/v) of a solution prepared by
dissolving 5.0 g of KI to give a final volume of 250 mL?
5.0 g
Mass % (m/v) =
x100 = 2.0 %
250 mL
7
USING PERCENT
CONCENTRATION

 Some
In theexamples
preparation
of percent
of solutions,
compositions,
one oftentheir
needs to
meanings,
calculate the
andamount
possible
ofconversion
solute or solution.
factors are
in thethis,
table
below:composition can be used as
 shown
To achieve
percent
a conversion factor.
8
Example 1:
A topical antibiotic solution is 1.0% (m/v) Clindamycin.
How many grams of Clindamycin are in 65 mL of this
solution?
1.0 g Clindamycin
65 mL solution x
100 mL solution = 0.65 g
9
Example 2:
How many grams of KCl are in 225 g of an 8.00% (m/m)
solution?
8.00 g KCl
225 g solution x
100 g solution = 18.0 g KCl
10
Example 3:
How many grams of solute are needed to prepare 150 mL
of a 40.0% (m/v) solution of LiNO3?
40.0 g LiNO3
150 mL solution x
100 mL solution = 60. g LiNO3
11
MOLARITY
 The most common unit of concentration used
in the laboratory is molarity (M).
 Molarity is defined as:
Molarity =
moles of solute
Liter of solution
12
Example 1:
What is the molarity of a solution containing 1.4 mol
of acetic acid in 250 mL of solution?
1L
= 0.25 L
Vol. of solution = 250 mL x
1000 mL
1.4 mol acetic acid
= 5.6 M
Molarity =
0.25 L
13
Example 2:
What is the molarity of a solution prepared by
dissolving 60.0 g of NaOH in 0.250 L of solution?
1 mol
= 1.50 mol
Mol of solute = 60.0 g x
40.0 g
1.50 mol
= 6.00 M
Molarity =
0.250 L
14
Example 3:
What is the molarity of a solution that contains 75 g
of KNO3 in 350 mL of solution?
1 mol
Mol of solute = 75 g x
= 0.74 mol
101.1 g
1L
= 0.35 L
Vol of solvent = 350 mL x
1000 mL
0.74 mol
Molarity =
= 2.1 M
0.350 L
15
USING
MOLARITY
 Molarity relationship can be used to calculate:
moles solute
Molarity =
volume of solution
Amount of solute:
Moles solute = Molarity x volume
Volume of solution:
moles solute
Volume of solution =
Molarity
16
Example 1:
How many moles of nitric acid are in 325 mL of 16 M
HNO3 solution?
1L
= 0.325 L
Vol. of solution = 325 mL x
1000 mL
mol of solute = 0.325 L x
16 mol
= 5.2 mol
1 L
17
Example 2:
How many grams of KCl would you need to prepare
0.250 L of 2.00 M KCl solution?
2.00 mol
mol of solute = 0.250 L x
= 0.500 mol
1 L
74.6 g
mass of solute = 0.500 mol x
= 37.3 g
1 mol
18
Example 3:
How many grams of NaHCO3 are in 325 mL of 4.50 M
solution of NaHCO3?
1L
= 0.325 L
Vol. of solution = 325 mL x
1000 mL
4.50 mol
mol of solute = 0.325 L x
= 1.46 mol
1 L
84.0 g
= 123 g
mass of solute = 1.46 mol x
1 mol
19
Example 4:
What volume (L) of 1.5 M HCl solution contains 6.0
moles of HCl?
1 L
Vol. of solution = 6.0 mol x
= 4.0 L
1.5 mol
20
Example 5:
What volume (mL) of 2.0 M NaOH solution contains
20.0 g of NaOH?
1 mol
mol of solute = 20.0 g x
=
0.500
mol
40.0 g
L
1
Vol. In L = 0.500 mol x
= 0.25 L
2.0 mol
1000 mL
Vol. In mL = 0.250 L x
= 250 mL
1L
21
Example 6:
How many mL of a 0.300 M glucose (C6H12O6) IV solution
is needed to deliver 10.0 g of glucose to the patient?
1 mol
mol of solute = 10.0 g x
= 0.0555 mol
180.1 g
L
1
Vol. In L = 0.0555 mol x
= 0.185 L
0.300 mol
1000 mL
Vol. In mL = 0.185 L x
= 185 mL
1L
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DILUTION
Amount of
solute

Solutions
When
arewater
oftenisprepared
added
tofrom
a solution,
more
Volume
andmore
concentration
are inversely
proportional
remains
concentrated
ones
by adding water. This
Volume
constant
process is called
dilution.
increases
Concentration
decreases
Frozen
juice
Water
Diluted
juice
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DILUTION
 The amount of solute depends on the
concentration and the volume of the solution.
Therefore,
M1 x V1 = M2 x V2
Concentrated
solution
Dilute
solution
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Example 1:
What is the molarity of the final solution when 75 mL of
Concentration
6.0 M KCl solution is diluted to 150 mL?
M1 = 6.0 M
V1 = 75 mL
decreases
Volume
increases
M1 x V1 = M2 x V2
M 1 V1 (6.0 M)(75 mL)
M2 =
=
V2
150 mL
M2 = ???
V2 = 150 mL
M2 = 3.0 M
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Example 2:
What volume (mL) of 0.20 M HCl solution can be
Volume
prepared by diluting 50.0
mL
of
1.0
M
HCl?
Concentration
decreases
M1 = 1.0 M
V1 = 50.0 mL
M2 = 0.20 M
V2 = ???
increases
M1 x V1 = M2 x V2
M 1 V1 (1.0 M)(50.0 mL)
V2 =
=
M2
0.20 M
V2 = 250 mL
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OSMOLARITY
 Many
Recall important
that when properties
ionic substances
of solutions
(strongdepend
on the number
electrolytes)
dissolve
of particles
in water
formed
they form
in solution.
several
particles for each formula unit.
 For example:
NaCl (s)
1 formula
unit
Na+ (aq) + Cl (aq)
2 particles
27
OSMOLARITY
CaCl2 (s)
1 formula
unit
Ca2+ (aq) + 2 Cl (aq)
3 particles
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OSMOLARITY
 When covalent substances (non- or weak
electrolytes) dissolve in water they form only
one particle for each formula unit.
 For example:
C12H22O11 (s)
1 formula
unit
C12H22O11 (aq)
1 particle
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OSMOLARITY
 Osmolarity of a solution is its molarity
multiplied by the number of particles formed
in solution.
Osmolarity =
i
x Molarity
Number of
particles in
solution
30
Examples:
0.10 M NaCl =
1 particle
2 x 0.10 M
= 0.20inosmol
solution
0.10 M CaCl2 = 3 x 0.10 M = 0.30 osmol
0.10 M C12H22O112 =
1
x
0.10
M
=
0.10
osmol
particles
3 particles
in solution
in solution
Same molarities
but different osmolarities
31
TONICITY OF
SOLUTIONS
 Because the cell membranes in biological systems
are semipermeable, particles of solute in solutions
can travel in and out of the membranes. This
process is called osmosis.
 The direction of the flow of solutions in or out of the
cell membranes is determined by the relative
osmolarity of the cell and the solution.
 The comparison of osmolarity of a solution with
those in body fluids determines the tonicity of a
solution.
32
ISOTONIC
SOLUTIONS
 Solutions with the same osmolarity as the cells
(0.30) are called isotonic.
 These solutions are called physiological
solutions and allow red blood cells to
retain their normal volume.
33
HYPOTONIC
SOLUTIONS
 Solutions with lower osmolarity than the cells are
called hypotonic.
 In these solutions, water flows into a
red blood cell, causing it to swell and
burst (hemolysis).
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HYPERTONIC
SOLUTIONS
 Solutions with greater osmolarity than the cells
are called hypertonic.
 In these solutions, water leaves the
red blood cells causing it to shrink
(crenation).
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Examples:
0.10 M NaCl = 0.20 osmol
hypotonic
0.10 M CaCl2 = 0.30 osmol
isotonic
0.10 M C12H22O11 = 0.10 osmol
hypotonic
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THE END
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