Concentration of Solution
Solvent
Solute
Concentration of Solution
Moles of solute
Mol
=
•Molarity (M) =
Liter of solution
L
•Parts ratio
amount of solute (g or ml)
2) or (106) or (109)
(10
= amount of solution (g or ml)
•Mole Fraction
•Molality
(c)= Total moles of solution
(m) =
Moles of solute
Moles of solute
Kilograms of solvent
Molarity
NaCl
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
 1molNaCl 
12.6 g NaCl 

58.44
gNaCl


=
 1L 
344 mL 
 solution
 1000mL 
= 0.627 M NaCl
Molarity
NaCl
Molarity
Example Problem 2
How many moles of NaCl are contained in 250.mL
of solution with a concentration of 1.25 M?
moles solute
M=
L solution
 1L 
250. mL 
 = 0.250 L solution
 1000mL 
Volume
x concentration
therefore the
solution contains
1.25 mol NaCl
1 L solution
=
moles solute
 1.25 mol NaCl 
0.250 L solution 
 = 0.313
 1 L solution 
mol NaCl
Molarity
NaCl
Molarity
Example Problem 3
What volume of solution will contain 15 g of NaCl
if the solution concentration is 0.75 M?
moles solute
M=
L solution
 1 mol NaCl 
15 g NaCl 
 = 0.257 mol
 58.44 g NaCl 
therefore the
solution contains
0.75 mol NaCl
1 L solution
moles solute ÷ concentration = volume solution
 1 L solution 
0.257 mol NaCl 
 = 0.34
 0.75 mol NaCl 
L solution
% Concentration
mass solute
• % (w/w) =
mass solution
x 100
mass solute
• % (w/v) =
volume solution
x 100
volume solute
• % (v/v) =
volume solution
x 100
Mass and volume units must match.
(g & mL)
or
(Kg & L)
% 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?
39.2 g
mass solute
 100 = 22.1 % w/v
 100
% (w/v) =
volume solution
177mL
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?
volume solute
 100
% (v/v) =
volume solution
3.2 L
 100
6.5L
= 49 % v/v
% Concentration
Example Problem 3
What volume of 1.85 %w/v solution is needed to
provide 5.7 g of solute?
1.85 g solute
% (w/v) =
100 mL solution
We know:
g solute and
We want to get:
g solute
mL solution
 100 mL solution 
5.7 g solute 

1.85
g
solute


g solute ÷ concentration
mL solution
= 310 mL Solution
=
volume solution
Parts per million/billion (ppm & ppb)
mass solute
• ppm =
× 106
volume solution
or
mg
= ppm
L
mass solute
9
×
10
• ppb =
volume solution
or
g
L
= ppb
AND
Mass and volume units must match.
(g & mL)
or
(Kg & L)
For very low
concentrations:
ng
parts per trillion
L
= ppt
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
g solute
ppm =
×106
mL solution
6.75 g
6
ppm =
×10
mL
2.5×106 L  1000
1 L 
ppm = 0.0027
or
mg solute
ppm =
L solution
ppm =
6.75 g 
1000 mg
1 g
2.5×106 L
ppm = 0.0027

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
g solute
ppb =
×109
mL solution
6.75 g
9
ppb =
×10
mL
2.5×106 L  1000
1 L 
ppb = 2.7
or
ppb =
 g solute
L solution
6.75 g
ppb =

106 mg
1 g
2.5×106 L
ppb = 2.7

Mole Fraction
B
A
A
B
A
A
B
A
A
B
A
B
A
A
Mole Fraction (c)
moles of A A
c A = sum of moles of all components
moles of B B
c B = sum of moles of all components
Since A + B make up the
entire mixture, their mole
fractions will add up to one.
A
A
+
+
B
B
c A  cB  1.00
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.
 1 mol C12 H 22 O11 
75.g C12 H 22O11 
 = 0.219 mol
342
g
C
H
O

12
22 11 
 1 mol H 2 O 
325mL H2O 
 = 18.1 mol
 18.0 g H 2 O 
Next, we substitute the moles of both into the mole fraction equation.
χ
sugar =
0.219 mol sugar
moles solute
=
(0.219 mol + 18.1 mol)
total moles solution
 0.012
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.
 1 mol N 2 
78g N 2 
 = 2.79 mol
 28 g N 2 
 1 mol O2 
21g O2 
 = 0.656 mol
32
g
O

2 
 1 mol Ar 
0.90g Ar 
 = 0.0225 mol
40.
g
A
r


χ
=
moles N 2
=
N2 total moles
2.79 mol N 2
(2.79 + 0.656 + 0.0225)
 0.804
Next, we substitute the moles of each into
the mole fraction equation.
χ
=
χ
moles O 2
=
O2 total moles
0.656 mol O2
(2.79 + 0.656 + 0.0225)
 0.189
=
=
Ar
moles Ar
total moles
0.0225 mol Ar
(2.79 + 0.656 + 0.0225)
 0.00649
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?
mol solute
m=
Kg solvent
m=
 1053 g C2 H6O2 
6 Qts 

1
Qt
C
H
O
2 6 2 

 946 g H 2O 
8 Qts 

1
Qt
H
O

2

antifreeze is ethylene glycol C2H6O2
1 qt antifreeze = 1053 grams
1 qt water = 946 grams
 1mol C2H6O2 


62.1
g
C
H
O
2 6 2 

 1 Kg 
 1000 g 


= 13 m