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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
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Concentration, continued Calculating Concentration, continued
• Concentrations can be expressed in many forms.
Section 2 Concentration and
Molarity Chapter 13
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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.
Section 2 Concentration and
Molarity Chapter 13
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Calculating Parts per Million, continued
Sample Problem A Solution
mass of solute: 2.2 mg
mass of solvent: 500 g
parts per million = ?
Section 2 Concentration and
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)
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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.
Section 2 Concentration and
Molarity Chapter 13
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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.
Section 2 Concentration and
Molarity Chapter 13
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Preparing 1.000 L of a 0.5000 M Solution
Chapter 13 Section 2 Concentration and
Molarity
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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.
Section 2 Concentration and
Molarity Chapter 13
moles of solutemolarity =
liters of solution
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Calculating Molarity Given Mass of Solute and Volume of Solution
Chapter 13 Section 2 Concentration and
Molarity
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Calculating Mass of Solute Given Molarity
and Volume of Solution
Chapter 13 Section 2 Concentration and
Molarity
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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?
Section 2 Concentration and
Molarity Chapter 13
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Calculating Molarity
Sample Problem B Solution
volume of solution = 400.0 mL
mass of solute = 85.0 g KCl
molarity of KCl solution = ?
Section 2 Concentration and
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
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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.
Section 2 Concentration and
Molarity Chapter 13
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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?
Section 2 Concentration and
Molarity Chapter 13
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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)
Section 2 Concentration and
Molarity Chapter 13
Solution Stoichiometry, continued
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Sample Problem B Solution, continued
3CuSO4(aq) + 2Al(s) → 3Cu(s) + Al2(SO4)3(aq)
Section 2 Concentration and
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
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