This document defines and provides examples of different types of concentration units used to quantify the amount of solute dissolved in a solution. It discusses molarity, percent concentration, parts per million/billion, mole fraction, and molality. Example problems are provided for each type of unit to demonstrate how to calculate concentration from given amounts of solute and solvent.

2. Concentration of Solution
Solvent Solute

3. •Molarity
•Parts ratio
•Mole Fraction
•Molality
Concentration of Solution
Moles of solute
Liter of solution
(M) =
=
Mol
L
amount of solute (g or ml)
amount of solution (g or ml)
(102) or (106) or (109)
Moles of solute
Total moles of solution
(c)
=
=
Kilograms of solvent
Moles of solute
(m) =

4. Molarity
Molarity Example Problem 1
12.6 g of NaCl are dissolved in water making
344mL of solution. Calculate the molar
concentration.
moles solute
M =
L solution
1
12.6 g NaCl
58.44
=
1
344 mL solution
1000
molNaCl
gNaCl
L
mL
 
 
 
 
 
 
= 0.627 M NaCl
NaCl

5. Molarity
Molarity Example Problem 2
How many moles of NaCl are contained in 250.mL
of solution with a concentration of 1.25 M?
therefore the
solution contains
1.25 mol NaCl
1 L solution
1
250. mL = 0.250 L solution
1000
L
mL
 
 
 
1.25 mol NaCl
0.250 L solution
1 L solution
 
 
 
NaCl
moles solute
M =
L solution
Volume x concentration = moles solute
= 0.313 mol NaCl

6. Molarity
Molarity Example Problem 3
What volume of solution will contain 15 g of NaCl
if the solution concentration is 0.75 M?
therefore the
solution contains
0.75 mol NaCl
1 L solution
1 mol NaCl
15 g NaCl = 0.257 mol
58.44 g NaCl
 
 
 
1 L solution
0.257
0.75 mol NaCl
mol NaCl
 
 
 
NaCl
moles solute
M =
L solution
moles solute ÷ concentration = volume solution
= L solution
0.34

7. • % (w/w) =
• % (w/v) =
• % (v/v) =
% Concentration
100
x
solution
mass
solute
mass
100
x
solution
volume
solute
mass
100
x
solution
volume
solute
volume
Mass and volume units must match.
(g & mL) or (Kg & L)

8. % Concentration
Example Problem 1
What is the concentration in %w/v of a solution containing 39.2 g
of potassium nitrate in 177 mL of solution?
100
mass solute
volume solution

% (w/v) =
39.2
100
177
g
mL
 = 22.1 % w/v
Example Problem 2
What is the concentration in %v/v of a solution containing 3.2 L of
ethanol in 6.5 L of solution?
100
volume solute
volume solution

% (v/v) =
3.2
100
6.5
L
L
 = 49 % v/v

9. % Concentration
Example Problem 3
What volume of 1.85 %w/v solution is needed to
provide 5.7 g of solute?
100 mL solution
5.7 g solute
1.85 g solute
 
 
 
% (w/v) =
1.85 g solute
100 mL solution
= 310 mL Solution
g solute ÷ concentration = volume solution
We know:
g solute
g solute and
mL solution
We want to get:
mL solution

10. • ppm =
• ppb =
Parts per million/billion (ppm & ppb)
6
mass solute
× 10
volume solution
Mass and volume units must match.
(g & mL) or (Kg & L)
9
mass solute
× 10
volume solution
or
or
mg
L
g
L

= ppm
= ppb
AND
For very low
concentrations:
ng
L
= ppt
parts per trillion

11. ppm & ppb
Example Problem 1
An Olympic sized swimming pool
contains 2,500,000 L of water. If 1 tsp of
salt (NaCl) is dissolved in the pool, what
is the concentration in ppm?
1 teaspoon = 6.75 g NaCl
6
g solute
ppm = ×10
mL solution
 
6
6 1000 mL
1 L
6.75 g
ppm = ×10
2.5×10 L
ppm = 0.0027
or
mg solute
ppm =
L solution
 
1000 mg
1 g
6
6.75 g
ppm =
2.5×10 L
ppm = 0.0027

12. ppm & ppb
Example Problem 2
An Olympic sized swimming pool
contains 2,500,000 L of water. If 1 tsp of
salt (NaCl) is dissolved in the pool, what
is the concentration in ppb?
1 teaspoon = 6.75 g NaCl
9
g solute
ppb = ×10
mL solution
 
9
6 1000 mL
1 L
6.75 g
ppb = ×10
2.5×10 L
ppb = 2.7
or
g solute
ppb =
L solution

 
6
10 mg
1 g
6
6.75 g
ppb =
2.5×10 L
ppb = 2.7

13. Mole Fraction
Mole Fraction (c)
A
B
B
B
B
B
A
A
A
A
A
A
A
A
c A =
moles of A
sum of moles of all components
A
B
A +
c B =
moles of B
sum of moles of all components
B
B
A +
Since A + B make up the
entire mixture, their mole
fractions will add up to one.
1.00
B
A
c c
 

14. Mole Fraction
Example Problem 1
In our glass of iced tea, we have added 3 tbsp
of sugar (C12H22O11). The volume of the tea
(water) is 325 mL. What is the mole fraction
of the sugar in the tea solution?
(1 tbsp sugar ≈ 25 g)
First, we find the moles of both the
solute and the solvent.
12 22 11
12 22 11
12 22 11
C H O
C H O
1 mol
75.g C H O =
342
0
g
.2 9 ol
1 m
 
 
 
2
2
2
H O
H O
1 mol
325mL H O =
18.0
18.1 m
g
ol
 
 
 
Next, we substitute the moles of both into the mole fraction equation.
sugar
moles solute
=
total moles solution
χ 0.219 mol sugar
=
(0.219 mol + 18.1 mol)
0.012


15. Mole Fraction
Example Problem 2
Air is about 78% N2, 21% O2, and 0.90% Ar.
What is the mole fraction of each gas?
First, we find the moles of each gas. We assume
100. grams total and change each % into grams.
2
2
2
1 mol N
78g N =
28 g
2.
N
79 mol
 
 
 
Next, we substitute the moles of each into
the mole fraction equation.
2
2
N
moles N
=
total moles
χ
2
(2.79 + 0.656 + 0.0225)
2.79 mol N
=
2
2
2
1 mol O
21g O =
32 g O
0.656 mol
 
 
 
1 mol Ar
0.90g Ar =
40. g
0.0225 m
A
ol
r
 
 
 
2
2
O
moles O
=
total moles
χ moles
=
total moles
χAr
Ar
2
(2.79 + 0.656 + 0.0225)
0.656 mol O
=
(2.79 + 0.656 + 0.0225)
0.0225 mol
=
Ar
0.804
 0.189
 0.00649


16. Molal (m)
Example Problem 1
If the cooling system in your
car has a capacity of 14 qts,
and you want the coolant to be protected from freezing
down to -25°F, the label says to combine 6 quarts of
antifreeze with 8 quarts of water. What is the molal
concentration of the antifreeze in the mixture?
antifreeze is ethylene glycol C2H6O2
1 qt antifreeze = 1053 grams
1 qt water = 946 grams
mol solute
m=
Kg solvent
2 6 2
2 6 2
1053 g C H O
6 Qts
1 Qt C H O
 
 
 
m =
2 6 2
2 6 2
1mol C H O
62.1 g C H O
 
 
 
2
2
946 g H O
8 Qts
1 Qt H O
 
 
 
1 Kg
1000 g
 
 
 
= 13 m