Concentration
ppm
Learning Goals
 Determine the concentration of a solution as a
percentage concentration (%m/m, %v/v, %m/v)
 Determine the concentration of very dilute solutions
as parts per million (ppm) parts per billion (ppb)
 Solve problems related to the concentration of a
solution by performing calculations and expressing the
concentration as a % or in ppm and ppb
Concentration of Solutions
 Concentration refers to the quantity of solute dissolved
in a given quantity of solvent
 Qualitatively, we describe solutions as either
concentrated or dilute
Concentration….
 Is the ratio of the quantity of solute to the quantity of
solution
 Concentration = quantity of solute

= C
quantity of solution
 Concentration can be expressed in many ways,
depending on the type of solution you have and what
you are using it for.
Ways of expressing concentration
of solutions
Percent concentration – on most consumer products
2. Parts per million (ppm) for very dilute solutions
3. Molar concentration – when doing stoichiometric
calculations.
1.
1. Percentage Concentration
 Usually used on consumer products
 Expressed as a percentage
 m/v used when a solid solute is dissolved in a solvent
 m/m used when 2 solids are mixed (alloys)
 v/v used when 2 liquids are mixed
Concentration - % M/V (%w/v)
 This concentration usually describes
a solid solute dissolved in a liquid
solvent.
%(m/v) 
mass of solute(g)
 100%
volume of solution (mL)
 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)
m/v concentrations:
 C solution =
𝑚 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 100%
 C solution = concentration of solution as % m/v
 𝑚 𝑠𝑜𝑙𝑢𝑡𝑒 = mass of solute in the solution (g)
 𝑣 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = volume of solution (mL)
Example 1
 An intravenous solution for a patient was
prepared by dissolving 17.5 g of glucose in
distilled water to make 350.0 mL of solution.
Find the % m/v concentration of the
solution.
Example 2
 A box of apple juice has a fructose concentration of of
12g/100mL (12% m/v). What mass of fructose is
present in a 175mL glass of juice?
Concentration - % M/M (%w/w)
 This concentration usually
describes a solid solute in a
solid solvent.
mass of solute (g)
%(m/m) 
 100%
mass of solution (g)
 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)
m/m concentrations
 𝑐 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 =
𝑚 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑥 100%
 𝑐 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = concentration of solution as % m/m
 𝑚 𝑠𝑜𝑙𝑢𝑡𝑒 = mass of solute
 𝑚 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = mass of solution
** both masses must have the same units**
Example 3
 Sterling silver is 92.5% silver and 7.5% copper. Find
the mass of pure silver in a sterling silver ring that has
a mass of 6.45g.
Example 4
 A sterling silver ring has a mass of 12.0g and contains
11.1g of pure silver. What is the percentage mass by
mass concentration of silver in the metal?
Concentration - % V/V (%v/v)
 This concentration usually describes
a liquid solute in a liquid solvent.
% v/v = volume of solute (mL) x 100%
volume of solution (mL)
 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.)
v/v concentrations
 𝑐 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 =
𝑣 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 100%
 𝑐 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = concentration of solution in % v/v
 𝑣 𝑠𝑜𝑙𝑢𝑡𝑒 = volume of solute
 𝑣 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = volume of solution
** both volumes must have the same units**
Example 5
 Acetic acid (CH3COOH) is a liquid at room
temperature. How much pure water shold be added to
15.0mL of pure acetic acid to make 5.00% v/v solution
of acetic acid? Assume that the total volume of the
solution equals the sum of the volumes of water and
pure acetic acid.
Example 6
 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 ethanol in a 55L tank of fuel.
Calculate the percentage by volume concentration of
ethanol.
2. Parts per million (ppm)
 Used to describe very dilute solutions
 Chemists also use parts per billion (ppb) or parts per
trillion (ppt)
 Uses the same ratio as the % m/m but multiplied by
106 for ppm and 109 for ppb
 𝑝𝑝𝑚 =
𝑚 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x
106
𝑝𝑝𝑏 =
𝑚 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 109
Expressing ppm in various units
1ppm =
1 g/106 mL
1 g/1000 L
1 mg/L
1 mg/kg
1 μg/g
*Choose the unit that matches the information given
in the example you are calculating.
Toxicity of hydrogen cyanide HCN
Example 7
 Health Canada’s guidelines for the maximum mercury
content in commercial fish is 0.5 ppm. When a 1.6kg
salmon was tested, it was found to contain0.6mg of
mercury. Would this salmon be safe to eat?
Example 8
 If the concentration of oxygen in water is 8 ppm, what
mass of oxygen is present in 150mL of water?
Molar Concentration - Molarity (M)
 This concentration unit describes the amount of solute (in
moles) dissolved in 1 litre of solution.
Molar concentration = amount of solute (moles)
volume of solution (litres)
C = n_
v
 Example of a solution that uses this concentration unit:
- Laboratory solutions - 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.
C. Molarity
Concentration of a solution most often
used by chemists
substance being dissolved
moles of solute
Molarity (C) 
liters of solution
total combined volume
C. Molarity
2M HCl
What does this mean?
mol
M
L
2 mol HCl
2M HCl 
1 L sol' n
Calculating Molarity
C = molar concentration
n = moles
V = volume in litres
D. Molarity Calculations
LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS
IN
GRAMS
Molar Mass
(g/mol)
NUMBER
6.02 
OF
particles/mol
PARTICLES
1023
MOLES
Molarity
(mol/L)
LITERS
OF
SOLUTION
D. Molarity Calculations
How many moles of NaCl are
required to make 0.500L of
0.25M NaCl?
x mol
0.25M 
0.500 L
= 0.13 mol NaCl
D. Molarity Calculations
How many grams of NaCl are
required to make 0.500L of
0.25M NaCl?
x mol
0.25M 
0.500 L
= 7.3 g NaCl
D. Molarity Calculations
Find the molarity of a 250 mL
solution containing 10.0 g of NaF.
= 0.95 M
NaF
mol
M
L
E. Dilution
Preparation of a desired solution by adding
water to a concentrate
Moles of solute remain the same
C 1V1  C 2V2
E. Dilution
What volume of 15.8M HNO3 is required to
make 250 mL of a 6.0M solution?
GIVEN:
C1 = 15.8M
V1 = ?
C2 = 6.0M
V2 = 250 mL
WORK:
C1 V1 = C2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3
H. Preparing Solutions
 250 mL of 6.0M HNO3
by dilution
95 mL of
15.8M HNO3
measure 95 mL
of 15.8M HNO3
• combine with water until
total volume is 250 mL
• Safety: “Do as you
oughtta, add the acid to
the watta!” or AA – add
acid!
250 mL
mark
water
for
safety