concentrations

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