Concentrations of Solutions
Amounts and Volumes
Objectives
• When you complete this presentation, you will
be able to
o Distinguish between solute, solvent, and solution.
o Define the concentration of a solution in terms of
molarity and percent composition.
o Calculate the molarity of a solution given the
volume of the solution and number of mols of a
solute
Introduction
• The concentration of a solution is the
measure of the amount of a solute is in a
solution.
• A dilute solution has a small amount of solute.
• A concentrated solution has a large amount of
solute.
• These terms are too qualitative - we need a
quantitative measurement system.
Molarity
• We can use the number of mols of a solute, n,
in a volume, V, of solution to give us a
quantitative measurement.
• We call this Molarity (M).
number mols of solute
Molarity =
volume of solution (L)
n
M=
V
Molarity
n
M=
V
• We can rearrange the formula to solve for
mols or volume.
n = M×V
n
V=
M
Molarity
Example 1:
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?
• V = 100 mL = 0.100 L
mNaCl
0.90 g
n=
=
= 0.015 mol
• mNaCl = 0.90 g
MNaCl
58.5 g/mol
• MNaCl = 58.5 g/mol
This value comes from
First,
we
write
down
n
0.015 mol
the
atomic
mass
Next,
we
need
toof Na
Now,
we
have
enough
M=
=
= 0.15 M
the
known
values.
V
0.100 L
(23.0
g/mol)
plus
the
calculate
the
number
information
to
atomic
of
Cl
of molsmass
ofthe
NaCl
in
calculate
molarity.
and
dothe
thecalculation.
calculation.
We
substitute
our known
……and
do
(35.5
g/mol).
solution.
values …
Molarity
Find the molarity of each of the following
solutions.
1. 0.200 mols of NaOH in a solution of 500 mL
0.400 M
2. 1.25 mols of CuCl2 in a solution of 3.40 L
0.368 M
3. 0.0352 mols of KCl in a solution of 25.0 mL
1.41 M
4. 14.0 mols of Li2CO3 in a solution of 5.25 L
2.67 M
Molarity
Example 2:
What is the mass of CuCl2, M = 134 g/mol, in 425 mL of a 2.50 M
solution of copper(II) chloride?
Next, we
we have
needenough
to
Now,
• V = 425 mL = 0.425 L
First,
we write
down
calculate
thetonumber
information
• M = 2.50 M
the
known
values.
of
mols
ofthe
CuCl
calculate
mass
2 in of
• MCuCl2 = 134 g/mol
solution.
CuCl
.
2
n = MV = (2.50 M)(0.425 L) = 1.06 mol
m = Mn = (134 g/mol)(1.06 mol)= 142 g
We
substitute
our
… and
then do in
the
known values …
calculation.
Molarity
Find the number of mols of solute in each of the
following solutions.
1. 4.25 L of 0.250 M CuSO4
1.06 mol CuSO4
2. 12.5 mL of 6.50 M HCl
3. 125 mL of 0.0450 M Na3PO4
4. 25.0 mL of 2.37 M AgNO3
0.0815 mol HCl
0.00563 mol Na3PO4
0.0593 mol AgNO3
Making Dilutions
• If we take a solution and add solvent to it, we
are diluting the solution.
o We have kept the same amount of solute in the
solution and only changed the volume.
• The number of mols before dilution is equal to
the number of mols after dilution.
n1 = n2
Making Dilutions
• n1 = n2
• If we expand this by using n = M×V, we get:
o M1×V1 = M2×V2
o This is called the “dilution formula.”
• Dilutions are used with “stock” solutions.
o We dilute the stock solutions to the concentration
we desire.
Making Dilutions
Example 3:
How many milliliters of aqueous 2.00 M MgSO4 solution must be
diluted with water to prepare 100 mL of aqueous 0.400 M
MgSO4?
…
and
do write
the
Then,
We
Next,
First,
substitute
we
we
rearrange
writeour
down
down
to
calculation.
• M1 = 2.00 M
solve
known
our
the known
dilution
for
values
V1.values.
equation.
…
• V1 = ? mL g
• M2 = 0.400 M
• V2 = 100 mL
M2V2
(0.400 M)(100 mL)
=
= 20.0 mL
M1V1 = M2V2 ⟹ V1 =
M1
2.00 M
Making Dilutions
How much stock solution is needed to prepare:
1. 5.00 L of 0.250 M HCl from 12.0 M HCl?
0.104 L = 104 mL
2. 500 mL of 0.100 M NaOH from 6.00 M NaOH?
0.00833 L = 8.33 mL
3. 125 mL of 0.0250 M HNO3 from 13.5 M HNO3?
0.000231 L = 0.231 mL
4. 25.0 mL of 0.0500 M AgNO3 from 0.400 M
AgNO3?
0.00313 L = 3.13 mL
Percent Solutions
• The concentration of a solution in percent can
be expressed in two ways:
o as the ratio of the volume of the solute to the
volume of the solution, %(v/v).
• %(v/v) = (V /V
solute
)(100%)
solution
o as the ratio of the mass of the solute to the mass
of the solution, %(m/m).
• %(m/m) = (m /m )(100%)
solute
solution
Percent Solutions
Example 4:
What is the percent by volume of isopropanol in
the final solution when 85 mL of isopropanol is
diluted to a volume of 250 mL with water?
Visopropanol = 85 mL
Vsolution = 250 mL
%(v/v) = (Visopropanol/Vsolution)(100%)
%(v/v) = (85 mL/250 mL)(100%) = 34%(v/v)
Percent Solutions
What is the %(v/v) of
1. 25.0 ml of ethyl alcohol diluted to 100 mL?
25.0%(v/v)
2. 12.5 mL of methyl alcohol diluted to 2.50 L?
0.500%(v/v)
3. 32.0 mL of molasses diluted to 425 mL?
7.53%(v/V)
4. 15.6 mL of ethyl alcohol diluted to 100. mL?
15.6%(v/V)
Summary
• The concentration of a solution is the measure
of the amount of a solute is in a solution.
• We can use the number of mols of a solute in
a liter of solution to give us a quantitative
measurement called molarity, M.
• If we take a solution and add solvent to it, we
are diluting the solution.
• n1 = n2 ➔ M1V1 = M2V2
Summary
• The concentration of a solution in percent can
be expressed in two ways:
• as the ratio of the volume of the solute to the
volume of the solution, %(v/v).
o %(v/v) = (V /V
solute
solution
)(100%)
• as the ratio of the mass of the solute to the
mass of the solution, %(m/m).
o %(m/m) = (m /m
solute
)(100%)
solution