concentration of solution solvent solute molarity parts ratio mole fraction molality concentration...

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