Concentration of Solution
Solvent Solute
•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 soluteTotal moles of solution()
=
=
Kilograms of solventMoles of solute(m) =
Molarity
Molarity Example Problem 1
12.6 g of NaCl are dissolved in water making 344mL of solution. Calculate the molar
concentration.
moles soluteM =
L solution
112.6 g NaCl
58.44 =
1344 mL solution
1000
molNaClgNaCl
LmL
= 0.627 M NaCl
NaCl
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 solution1
250. mL = 0.250 L solution 1000
L
mL
1.25 mol NaCl0.250 L solution
1 L solution
NaCl
moles soluteM =
L solution
Volume x concentration = moles solute
= 0.313 mol NaCl
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 solution1 mol NaCl
15 g NaCl = 0.257 mol 58.44 g NaCl
1 L solution0.257
0.75 mol NaClmol NaCl
NaCl
moles soluteM =
L solution
moles solute ÷ concentration = volume solution
= L solution0.34
• % (w/w) =
• % (w/v) =
• % (v/v) =
% Concentration
100xsolutionmasssolutemass
100xsolutionvolumesolutemass
100xsolutionvolumesolutevolume
Mass and volume units must match.
(g & mL) or (Kg & L)
% ConcentrationExample Problem 1
What is the concentration in %w/v of a solution containing 39.2 g of potassium nitrate in 177 mL of solution?
100mass solute
volume solution% (w/v) =
39.2100
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?
100volume solute
volume solution% (v/v) =
3.2100
6.5
L
L = 49 % v/v
% ConcentrationExample Problem 3
What volume of 1.85 %w/v solution is needed to provide 5.7 g of solute?
100 mL solution5.7 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 soluteg solute and
mL solution
We want to get:
mL solution
• ppm =
• ppb =
Parts per million/billion (ppm & ppb)
6mass solute× 10
volume solution
Mass and volume units must match.
(g & mL) or (Kg & L)
9mass solute× 10
volume solution
or
or
mg
L
g
L
= ppm
= ppb
AND
For very low concentrations:
ng
L= pptparts per trillion
ppm & ppbExample 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
6g soluteppm = ×10
mL solution
6
6 1000 mL1 L
6.75 gppm = ×10
2.5×10 L
ppm = 0.0027
ormg solute
ppm = L solution
1000 mg1 g
6
6.75 gppm =
2.5×10 L
ppm = 0.0027
ppm & ppbExample 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
9g soluteppb = ×10
mL solution
9
6 1000 mL1 L
6.75 gppb = ×10
2.5×10 L
ppb = 2.7
org solute
ppb = L solution
610 mg1 g
6
6.75 gppb =
2.5×10 L
ppb = 2.7
Mole Fraction
Mole Fraction ()
A
B
B
B
B
B
A
A
A
A
A
A
A
A
A = moles of A
sum of moles of all components
A
B
A +
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.
1.00BA
Mole FractionExample Problem 1In 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 =
3420
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.
sugarmoles solute
=total moles solution
χ 0.219 mol sugar=
(0.219 mol + 18.1 mol) 0.012
Mole FractionExample 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.
22
2
1 mol N78g N =
28 g 2.
N79 mol
Next, we substitute the moles of each into the mole fraction equation.
2
2Nmoles N
=total moles
χ2
(2.79 + 0.656 + 0.0225)
2.79 mol N=
22
2
1 mol O21g O =
32 g O0.656 mol
1 mol Ar0.90g Ar =
40. g 0.0225 m
Aol
r
2
2Omoles O
=total moles
χ moles =
total moles χ Ar
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
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 grams1 qt water = 946 grams
mol solutem=
Kg solvent
2 6 2
2 6 2
1053 g C H O6 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 O8 Qts
1 Qt H O
1 Kg
1000 g
= 13 m