Transcript Ch16 Lesson16_2

16.2 Concentrations of Solutions >
Chapter 16
Solutions
16.1 Properties of Solutions
16.2 Concentrations of Solutions
16.3 Colligative Properties
of Solutions
16.4 Calculations Involving
Colligative Properties
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16.2 Concentrations of Solutions >
CHEMISTRY
& YOU
How can you describe the concentration
of a solution?
The federal government
and state governments
set standards limiting
the amount of
contaminants allowed in
drinking water.
2
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16.2 Concentrations of Solutions > Molarity
Molarity
How do you calculate the molarity of
a solution?
3
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16.2 Concentrations of Solutions > Molarity
The concentration of a solution is a
measure of the amount of solute that is
dissolved in a given quantity of solvent.
• A solution that contains a relatively small
amount of solute is a dilute solution.
• A concentrated solution contains a large
amount of solute.
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16.2 Concentrations of Solutions > Molarity
In chemistry, the most important unit of
concentration is molarity.
• Molarity (M) is the number of moles of
solute dissolved in one liter of solution.
• Molarity is also known as molar
concentration.
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16.2 Concentrations of Solutions > Molarity
The figure below illustrates the procedure
for making a 0.5M, or 0.5-molar, solution.
Add 0.5 mol of solute
to a 1-L volumetric
flask half filled with
distilled water.
6
Swirl the flask
carefully to
dissolve the
solute.
Fill the flask with
water exactly to
the 1-L mark.
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16.2 Concentrations of Solutions > Molarity
To calculate the molarity of a solution,
divide the number of moles of solute
by the volume of the solution in liters.
moles of solute
Molarity (M) =
liters of solution
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16.2 Concentrations of Solutions >
Sample Problem 16.2
Calculating Molarity
Intravenous (IV) saline solutions are often
administered to patients in the hospital.
One saline solution contains 0.90 g NaCl in
exactly 100 mL of solution. What is the
molarity of the solution?
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16.2 Concentrations of Solutions >
Sample Problem 16.2
1 Analyze List the knowns and the unknown.
Convert the concentration from g/100 mL
to mol/L. The sequence is:
g/100 mL → mol/100 mL → mol/L
KNOWNS
solution concentration = 0.90 g NaCl/100 mL
molar mass NaCl = 58.5 g/mol
UNKNOWN
solution concentration = ?M
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16.2 Concentrations of Solutions >
Sample Problem 16.2
2 Calculate Solve for the unknown.
Use the molar mass to convert
g NaCl/100 mL to mol NaCl/100 mL.
Then convert the volume units so
that your answer is expressed in
mol/L.
The relationship
1 L = 1000 mL
gives you the
conversion factor
1000 mL/1 L.
Solution
0.90 g NaCl
1 mol NaCl
1000 mL
 58.5 g NaCl 
concentration = 100 mL
1L
= 0.15 mol/L
= 0.15M
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16.2 Concentrations of Solutions >
Sample Problem 16.2
3 Evaluate Does the result make sense?
• The answer should be less than 1M because
a concentration of 0.90 g/100 mL is the same
as 9.0 g/1000 mL (9.0 g/1 L), and 9.0 g is less
than 1 mol NaCl.
• The answer is correctly expressed to two
significant figures.
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16.2 Concentrations of Solutions >
Sample Problem 16.3
Calculating the Moles of Solute
in a Solution
Household laundry bleach is a dilute
aqueous solution of sodium
hypochlorite (NaClO). How many
moles of solute are present in 1.5 L
of 0.70M NaClO?
12
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16.2 Concentrations of Solutions >
Sample Problem 16.3
1 Analyze List the knowns and the unknown.
The conversion is: volume of solution → moles
of solute. Molarity has the units mol/L, so you
can use it as a conversion factor between
moles of solute and volume of solution.
KNOWNS
volume of solution = 1.5 L
solution concentration = 0.70M NaClO
UNKNOWN
moles solute = ? mol
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16.2 Concentrations of Solutions >
Sample Problem 16.3
2 Calculate Solve for the unknown.
Multiply the given
volume by the
molarity expressed
in mol/L.
Make sure that your volume units
cancel when you do these
problems. If they don’t, then you’re
probably missing a conversion
factor in your calculations.
0.70 mol NaCl
1.5 L 
= 1.1 mol NaClO
1L
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16.2 Concentrations of Solutions >
Sample Problem 16.3
3 Evaluate Does the result make sense?
• The answer should be greater than 1 mol but
less than 1.5 mol, because the solution
concentration is less than 0.75 mol/L and the
volume is less than 2 L.
• The answer is correctly expressed to two
significant figures.
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16.2 Concentrations of Solutions >
How much water is required to make
a 1.00M aqueous solution of NaCl, if
58.4 g of NaCl are dissolved?
A. 1.00 liter of water
B. enough water to make 1.00 liter of
solution
C. 1.00 kg of water
D. 100 mL of water
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16.2 Concentrations of Solutions >
How much water is required to make
a 1.00M aqueous solution of NaCl, if
58.4 g of NaCl are dissolved?
A. 1.00 liter of water
B. enough water to make 1.00 liter of
solution
C. 1.00 kg of water
D. 100 mL of water
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16.2 Concentrations of Solutions > Making Dilutions
Making Dilutions
What effect does dilution have on
the amount of solute?
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16.2 Concentrations of Solutions > Making Dilutions
Both of these solutions contain the same
amount of solute.
• You can tell by the color of solution (a) that it is
more concentrated than solution (b).
• Solution (a) has
the greater
molarity.
• The more dilute
solution (b) was
made from
solution (a) by
adding more solvent.
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16.2 Concentrations of Solutions > Making Dilutions
Diluting a solution reduces the number
of moles of solute per unit volume, but
the total number of moles of solute in
solution does not change.
Moles of solute
Moles of solute
=
before dilution
after dilution
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16.2 Concentrations of Solutions > Making Dilutions
Moles of solute
Moles of solute
=
before dilution
after dilution
The definition of molarity can be rearranged
to solve for moles of solute.
moles of solute
Molarity (M) = liters of solution (V)
Moles of solute = molarity (M)  liters of solution (V)
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16.2 Concentrations of Solutions > Making Dilutions
The total number of moles of solute remains
unchanged upon dilution.
Moles of solute = M1  V1 = M2  V2
• M1 and V1 are the molarity and the volume of the initial
solution.
• M2 and V2 are the molarity and volume of the diluted
solution.
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16.2 Concentrations of Solutions > Making Dilutions
The student is preparing 100 mL of 0.40M MgSO4
from a stock solution of 2.0M MgSO4.
She measures 20 mL She transfers the
of the stock solution
20 mL to a 100-mL
with a 20-mL pipet.
volumetric flask.
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She carefully adds
water to the mark
to make 100 mL of
solution.
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16.2 Concentrations of Solutions >
Sample Problem 16.4
Preparing a Dilute Solution
How many milliliters of
aqueous 2.00M MgSO4
solution must be diluted with
water to prepare 100.0 mL of
aqueous 0.400M MgSO4?
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16.2 Concentrations of Solutions >
Sample Problem 16.4
1 Analyze List the knowns and the unknown.
Use the equation M1  V1 = M2  V2 to solve for
the unknown initial volume of solution (V1) that
is diluted with water.
KNOWNS
M1 = 2.00M MgSO4
M2 = 0.400M MgSO4
V2 = 100.0 mL of 0.400M MgSO4
UNKNOWN
V1 = ? mL of 2.00M MgSO4
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16.2 Concentrations of Solutions >
Sample Problem 16.4
2 Calculate Solve for the unknown.
Solve for V1 and substitute the known
values into the equation.
M2  V2
0.400M  100.0 mL
V1 =
=
= 20.0 mL
M1
2.00M
Thus, 20.0 mL of the initial solution must be
diluted by adding enough water to increase the
volume to 100.0 mL.
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16.2 Concentrations of Solutions >
Sample Problem 16.4
3 Evaluate Does the result make sense?
• The initial concentration is five times larger
than the dilute concentration.
• Because the number of moles of solute does
not change, the initial volume of solution
should be one-fifth the final volume of the
diluted solution.
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16.2 Concentrations of Solutions >
100.0 mL of a 0.300M CuSO4·5H2O
solution is diluted to 500.0 mL.
What is the concentration of the
diluted solution?
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16.2 Concentrations of Solutions >
100.0 mL of a 0.300M CuSO4·5H2O
solution is diluted to 500.0 mL.
What is the concentration of the
diluted solution?
M1  V1 = M2  V2
M1  V1 0.300M  100.0 mL
M2 =
=
V2
500.0 mL
M2 = 0.0600M
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16.2 Concentrations of Solutions > Percent Solutions
Percent Solutions
How do percent by volume and
percent by mass differ?
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16.2 Concentrations of Solutions > Percent Solutions
Percent by volume of a solution is
the ratio of the volume of solute to
the volume of solution.
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16.2 Concentrations of Solutions > Percent Solutions
Percent by volume of a solution is
the ratio of the volume of solute to
the volume of solution.
• Isopropyl alcohol (2-propanol) is sold
as a 91-percent solution by volume.
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16.2 Concentrations of Solutions > Percent Solutions
Percent by volume of a solution is
the ratio of the volume of solute to
the volume of solution.
• Isopropyl alcohol (2-propanol) is sold
as a 91-percent solution by volume.
• You could prepare such a solution by
diluting 91 mL of pure isopropyl
alcohol with enough water to make
100 mL of solution.
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16.2 Concentrations of Solutions > Percent Solutions
Percent by volume of a solution is
the ratio of the volume of solute to
the volume of solution.
• Isopropyl alcohol (2-propanol) is sold
as a 91-percent solution by volume.
• The concentration is written as 91
percent by volume, 91 percent
(volume/volume), or 91% (v/v).
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16.2 Concentrations of Solutions > Percent Solutions
Percent by volume of a solution is
the ratio of the volume of solute to
the volume of solution.
volume of solute
Percent by volume (%(v/v)) =
 100%
volume of solution
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16.2 Concentrations of Solutions >
Sample Problem 16.5
Calculating Percent by Volume
What is the percent by volume of ethanol
(C2H6O, or ethyl alcohol) in the final
solution when 85 mL of ethanol is diluted
to a volume of 250 mL with water?
36
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16.2 Concentrations of Solutions >
Sample Problem 16.5
1 Analyze List the knowns and the unknown.
Use the known values for the volume of solute
and volume of solution to calculate percent by
volume.
KNOWNS
volume of solute = 85 mL ethanol
volume of solution = 250 mL
UNKNOWN
Percent by volume = ?% ethanol (v/v)
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16.2 Concentrations of Solutions >
Sample Problem 16.5
2 Calculate Solve for the unknown.
State the equation for percent by
volume.
Percent by volume (%(v/v)) =
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volume of solute
volume of solution
 100%
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16.2 Concentrations of Solutions >
Sample Problem 16.5
2 Calculate Solve for the unknown.
Substitute the known values into the
equation and solve.
85 mL ethanol
Percent by volume (%(v/v)) =
 100%
250 mL
= 34% ethanol (v/v)
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16.2 Concentrations of Solutions >
Sample Problem 16.5
3 Evaluate Does the result make sense?
• The volume of the solute is about one-third
the volume of the solution, so the answer is
reasonable.
• The answer is correctly expressed to two
significant figures.
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16.2 Concentrations of Solutions > Percent Solutions
Another way to express the concentration
of a solution is as a percent by mass, or
percent (mass/mass).
Percent by mass of a solution is the ratio
of the mass of the solute to the mass of
the solution.
Percent by mass (%(m/m)) =
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mass of solute
mass of solution
 100%
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16.2 Concentrations of Solutions > Percent Solutions
mass of solute
Percent by mass (%(m/m)) =
 100%
mass of solution
• Percent by mass is sometimes a convenient
measure of concentration when the solute is a
solid.
• You have probably seen information on food
labels expressed as a percent composition.
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16.2 Concentrations of Solutions >
CHEMISTRY
& YOU
What are three ways to calculate the
concentration of a solution?
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16.2 Concentrations of Solutions >
CHEMISTRY
& YOU
What are three ways to calculate the
concentration of a solution?
The concentration
of a solution can be
calculated in moles
solute per liter of
solvent, or molarity
(M), percent by
volume (%(v/v)), or
percent by mass
(%(m/m)).
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16.2 Concentrations of Solutions >
Sample Problem 16.6
Using Percent by Mass as a
Conversion Factor
How many grams of
glucose (C6H12O6) are
needed to make 2000 g of
a 2.8% glucose (m/m)
solution?
45
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16.2 Concentrations of Solutions >
Sample Problem 16.6
1 Analyze List the knowns and the unknown.
The conversion is mass of solution → mass of
solute. In a 2.8% C6H12O6 (m/m) solution, each
100 g of solution contains 2.8 g of glucose. Used
as a conversion factor, the concentration allows
you to convert g of solution to g of C6H12O6.
KNOWNS
mass of solution = 2000 g
percent by mass = 2.8% C6H12O6(m/m)
UNKNOWN
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mass of solute = ? g C6H12O6
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16.2 Concentrations of Solutions >
Sample Problem 16.6
2 Calculate Solve for the unknown.
Write percent by mass as a conversion
factor with g C6H12O6 in the numerator.
2.8 g C6H12O6
100 g solution
47
You can solve this
problem by using
either dimensional
analysis or algebra.
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16.2 Concentrations of Solutions >
Sample Problem 16.6
2 Calculate Solve for the unknown.
Multiply the mass of the solution by the
conversion factor.
2000 g solution 
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2.8 g C6H12O6
100 g solution
= 56 g C6H12O6
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16.2 Concentrations of Solutions >
Sample Problem 16.6
3 Evaluate Does the result make sense?
• The prepared mass of the solution is 20  100 g.
• Since a 100-g sample of 2.8% (m/m) solution
contains 2.8 g of solute, you need 20  2.8 g = 56 g
of solute.
• To make the solution, mix 56 g of C6H12O6 with
1944 g of solvent.
• 56 g of solute + 1944 g solvent = 2000 g of solution
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16.2 Concentrations of Solutions >
What is the mass of water in a 2000 g
glucose (C6H12O6) solution that is
labeled 5.0% (m/m)?
50
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16.2 Concentrations of Solutions >
What is the mass of water in a 2000 g
glucose (C6H12O6) solution that is
labeled 5.0% (m/m)?
% (m/m) =
mass of glucose
 100%
mass of solution
(% (m/m))  mass of solution
mass of glucose =
100%
mass of glucose = 2000 g  0.050 = 100 g C6H12O6
mass of water = 2000 g – 100 g = 1900 g H2O
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16.2 Concentrations of Solutions > Key Concepts
To calculate the molarity of a solution,
divide the moles of solute by the volume
of the solution in liters.
Diluting a solution reduces the number of
moles of solute per unit volume, but the
total number of moles of solute in solution
does not change.
Percent by volume is the ratio of the
volume of solute to the volume of solution.
Percent by mass is the ratio of the mass of
the solute to the mass of the solution.
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16.2 Concentrations of Solutions > Key Equations
moles of solute
Molarity (M) =
liters of solution
M1  V1 = M2  V2
volume of solute
Percent by volume =
 100%
volume of solution
mass of solute
Percent by mass =
 100%
mass of solution
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16.2 Concentrations of Solutions > Glossary Terms
• concentration: a measurement of the amount
of solute that is dissolved in a given quantity of
solvent; usually expressed as mol/L
• dilute solution: a solution that contains a
small amount of solute
• concentrated solution: a solution containing
a large amount of solute
• molarity (M): the concentration of solute in a
solution expressed as the number of moles of
solute dissolved in 1 liter of solution
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16.2 Concentrations of Solutions >
END OF 16.2
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