Section 2 Concentration and - Crestwood Local Schools

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Concentration Calculation Concentration

• In a solution, the solute is distributed evenly

throughout the solvent. This means that any part of a

solution has the same ratio of solute to solvent as any

other part of the solution.

• This ratio is the concentration of the solution.

• The concentration is the amount of a particular

substance in a given quantity of a solution

Section 2 Concentration and

Molarity Chapter 13



Concentration, continued Calculating Concentration, continued

• Concentrations can be expressed in many forms.


Molarity Chapter 13



Calculating Parts per Million

Sample Problem A

A chemical analysis shows that there are

2.2 mg of lead in exactly 500 g of water. Convert this

measurement to parts per million.


Molarity Chapter 13



Calculating Parts per Million, continued

Sample Problem A Solution

mass of solute: 2.2 mg

mass of solvent: 500 g

parts per million = ?


Molarity Chapter 13

31 g2.2 mg 2.2 10 g

1000 mg

2

0.0022 g Pb 1000 000 parts

500 g H O 1 million

2 4.4 ppm (parts Pb per million parts H O)



Molarity

• Since the mole is the unit chemists use to measure

the number of particles, they often specify

concentrations using molarity.

• Molarity (M) is a concentration unit of a solution

expressed as moles of solute dissolved per liter of

solution.


Molarity Chapter 13



Molarity, continued Preparing a Solution of Specified Molarity

• Note that molarity describes concentration in terms of

volume of solution, not volume of solvent.

• If you simply added 1.000 mol solute to 1.000 L

solvent, the solution would not be 1.000 M.

• The added solute will change the volume, so the

solution would not have a concentration of 1.000 M.

• The solution must be made to have exactly the

specified volume of solution.


Molarity Chapter 13



Preparing 1.000 L of a 0.5000 M Solution

Chapter 13 Section 2 Concentration and

Molarity



Molarity, continued Calculating Molarity

• In working with solutions in chemistry, you will find

that numerical calculations often involve molarity.

• The key to all such calculations is the definition of

molarity, which is stated as an equation below.


Molarity Chapter 13

moles of solutemolarity =

liters of solution



Calculating Molarity Given Mass of Solute and Volume of Solution


Molarity



Calculating Mass of Solute Given Molarity

and Volume of Solution


Molarity



Calculating Molarity

Sample Problem B

What is the molarity of a potassium chloride solution

that has a volume of 400.0 mL and contains 85.0 g

KCl?


Molarity Chapter 13



Calculating Molarity

Sample Problem B Solution

volume of solution = 400.0 mL

mass of solute = 85.0 g KCl

molarity of KCl solution = ?


Molarity Chapter 13

1 L400.0 mL 0.4000 L

1000 mL

1.14 mol KCl2.85 mol/L 2.

0.400085 M

L KCl

1 mol85.0 g KCl 1.14 mol KCl

74.55 g KCl



Molarity, continued Using Molarity in Stoichiometric Calculations

• There are many instances in which solutions of

known molarity are used in chemical reactions in the

laboratory.

• Instead of starting with a known mass of reactants

or with a desired mass of product, the process

involves a solution of known molarity.

• The substances are measured out by volume,

instead of being weighed on a balance.


Molarity Chapter 13



Solution Stoichiometry

Sample Problem C

What volume (in milliliters) of a 0.500 M solution of

copper(II) sulfate, CuSO4, is needed to react with an

excess of aluminum to provide 11.0 g of copper?


Molarity Chapter 13



Sample Problem B Solution

[CuSO4] = 0.500 M

mass of product = 11.0 g Cu

solution volume = ? L

molar mass of Cu = 63.55 g /mol

3CuSO4(aq) + 2Al(s) → 3Cu(s) + Al2(SO4)3(aq)


Molarity Chapter 13

Solution Stoichiometry, continued



Sample Problem B Solution, continued

3CuSO4(aq) + 2Al(s) → 3Cu(s) + Al2(SO4)3(aq)


Molarity Chapter 13

4

4

3 mol CuSO1 mol Cu11.0 g Cu

63.55 g Cu 3 mol Cu

1 L solution 1000 mL solution

0.500 mol CuSO 1 L solution

4

346 mL CuSO solution

Solution Stoichiometry, continued

Section 2 Concentration and - Crestwood Local Schools

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