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