Unit 4: Solutions and Solubility
6.3 Solution Concentration
What is “Solution Concentration”?
Concentration = the quantity of a given
solute in a solution
In general,
Concentration = quantity of solute
quantity of solution
Some Important Terms
Dilute = having a relatively small quantity
of solute per unit volume of solution
Concentrated = having a relatively large
quantity of solute per unit volume of
solution
Concentration Units
% W/V (%m/v)
 This concentration usually describes a solid
solute dissolved in a liquid solvent.
 Example of a solution that uses this
concentration unit:
- Intravenous solution is 0.90% (m/v) NaCl
(this means 0.90g NaCl dissolved in 100mL of
solution)
Example 1
 A box of apple juice has a fructose (sugar)
concentration of 12 g/100 mL (12% W/V). What mass of
fructose is present in a 175 mL glass of juice)?
 volume of solution (apple juice) = 175 mL
% W/V = 12 %
 mass of solute (fructose) = ?
 %W/V
= mass of solute X 100
volume of solution
mass of solute
= % W/V X volume of solution
100
= 12 g/mL X 175 mL
100
= 21 g
Example 1
 A box of apple juice has a fructose (sugar)
concentration of 12 g/100 mL (12% W/V). What mass of
fructose is present in a 175 mL glass of juice)?
OR
Mass of fructose
= 12 g
X
100 mL
= 21 g
175 mL
% W/W (%m/m)
 This concentration usually
describes a solid solute in a solid
solvent.
%(m / m) =
mass of solute
mass of solution
X 100%
 Example of a solution that uses this
concentration unit:
- Toothpaste is
0.24% (m/m) SnF2
(this means 0.24g SnF2 dissolved in100g of
solution)
Example 2
 A sterling silver ring has a mass of 12.0 g and
contains 11.1 g of pure silver. What is the
percentage weight by weight concentration of
silver in the metal?
 mass of solute = 11.1 g
mass of solution = 12.0 g
 % W/W = ?
 % W/W = mass of solute X
mass of solution
= 11.1 g
X
12.0 g
= 92.5 %
100
100
% V/V (%v/v)
 This concentration usually describes a liquid
solute in a liquid solvent.
% v/v = volume of solute
x 100%
volume of solution
 Example of a solution that uses this
concentration unit:
- Wine is
11.0% (v/v) Ethanol (C2H5OH)
(this means 11.0 mL of ethanol is dissolved in 100mL
of solution, ie the wine.)
Example 3
 Gasohol, which is a solution of ethanol and gasoline, is considered to be
a cleaner fuel than just gasoline alone. A typical gasohol mixture available
across Canada contains 4.1L of ethanol in a 55L tank of fuel. Calculate
the percentage by volume concentration of ethanol.
% v/v
= volume of ethanol
volume of solution
= 4.1L X 100
55L
= 7.5%
X
100
parts per million - ppm
 This concentration unit is for very small
quantities of solute
 c (ppm) = amount of solute
amount of solution
 Example of a solution that uses this
concentration unit:
- Bottled water contains
280 ppm of HCO31-, 118 ppm Ca…
(this means every 1 000 000 parts of solution
contain 280 parts of HCO31- solute)
*1ppm = 1 drop of water in a bathtub
Units for ppm
1ppm =
1 mg/L
1 mg/kg
1 μg/g
*Choose the unit that matches the information
given in the example you are calculating.
Example 4:
If the concentration of oxygen in water is 8 ppm,
what mass of oxygen is present in 150 mL of
water?
c = 8 ppm (mg/L)
v (volume of solution) = 150 mL = 0.15 L
m (mass of oxygen) = ?
c
=
m
v
m
=
c
m
=
8 mg/L
m
=
1 mg
X
v
X
0.15 L
Molar Concentration - Molarity (M)
 This concentration unit describes the amount of
solute (in moles) dissolved in 1 L of solution.
Molar concentration = amount of solute (mol)
volume of solution (L)
C=n
v
 Example of a solution that uses this concentration
unit:
- Hydrochloric acid - HCl 1.25M (mol/L)
This means that 1.25 moles of HCl are dissolved in
1.00L of solution.
Chemists use Molarity because it allows them to relate
concentration to moles of a substance.
Example 5:
When 2.00 g of KMnO4 is dissolved into 100.0
mL of solution, what molar concentration
results?
m = 2.00 g
V = 100.0 mL
C=?
C
n
=
=
n
=
=
n
V
m
M
C
=
=
2.00 g
158.04g/mol
0.0127 mol
0.0127 mol
0.1000 L
0.127 mol/L
Example 6:
How many grams of KMnO4 are needed to
make 500.0 mL of a 0.200 mol/L solution?
C = 0.200 mol/L
V = 500.0 mL (0.5000L)
m=?
C
=
n
=
=
=
m
= M
X
n
= 158.04 g/mol X 0.100 mol
n
= 15.8 g
V
C
X
V
0.200 mol/L X 0.5000L
0.100 mol