Section 8.6: Concentration
Tutorial 1 Practice, page 400–401
1. Given: volume of calcium chloride, VCaCl = 750 mL
2
amount of calcium chloride, nCaCl = 1.5 mol
2
Required: amount concentration of calcium chloride solution, cCaCl
Analysis: cCaCl =
2
nCaCl
2
VCaCl
2
2
Solution:
Step 1. Convert the volume of solution to litres.
1L
VCaCl = 750 mL !
2
1000 mL
VCaCl = 0.75 L
2
Step 2. Substitute the values into the equation and solve.
1.5 mol
cCaCl =
2
0.75 L
mol
cCaCl = 2.0
2
L
Statement: The amount concentration of calcium chloride solution is 2.0 mol/L.
2. Given: mass of phosphoric acid, mH PO = 63 g
3
4
volume of phosphoric acid, VH PO = 250 mL
3
4
Required: amount concentration of phosphoric acid solution, cH PO
3
Analysis: cH PO =
3
4
nH PO
4
VH PO
4
3
3
4
Solution:
Step 1. Convert the volume of solution to litres.
1L
VH PO = 250 mL !
3
4
1000 mL
VH PO = 0.25 L
3
4
Step 2. Convert the mass of solute to amount of solute.
1 mol
nH PO = 63 g !
3
4
98.00 g
nH PO = 0.6428 mol [2 extra digits carried]
3
4
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Chapter 8: Water and Solutions
8.6-1
Step 3. Substitute the values into the equation and solve.
0.6428 mol
cH PO =
3
4
0.25 L
mol
cH PO = 2.6
3
4
L
Statement: The amount concentration of phosphoric acid solution is 2.6 mol/L.
3. Given: concentration of fructose, cC H O = 0.67 mol/L
6
volume of fructose, VC H
6
12 O6
6
6
12 O6
=
nC H
12 O6
VC H
12 O6
6
6
6
= 250 mL
Required: mass of fructose, mC H
Analysis: cC H
12
12 O6
Solution:
Step 1. Convert the volume of solution to litres.
1L
VC H O = 250 mL !
6 12 6
1000 mL
VC H O = 0.250 L
6
12
6
Step 2. Rearrange the equation and substitute values.
nC H O = cC H O VC H O
6
12
6
6
=
nC H
6
12 O6
12
6
6
12
6
0.67 mol
! 0.250 L
1L
= 0.1675 mol (2 extra digits carried)
Step 3. Convert the amount of solute to mass.
180.16 g
mC H O = 0.1675 mol !
6 12 6
1 mol
mC H O = 30 g
6
12
6
Statement: The mass of fructose in 450 mL of a 0.67 mol/L solution is 30 g.
4. Given: concentration of sodium hydroxide, cNaOH = 6.0 mol/L
mass of sodium hydroxide, mNaOH = 125 g
Required: volume of sodium hydroxide solution, VNaOH
Analysis: cNaOH =
nNaOH
VNaOH
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Chapter 8: Water and Solutions
8.6-2
Solution:
Step 1. Convert the mass of solute to amount of solute.
1 mol
nNaOH = 125 g !
40.00 g
nNaOH = 3.1250 mol [2 extra digits carried]
Step 2. Rearrange the equation and substitute values.
n
VNaOH = NaOH
cNaOH
=
3.1250 mol
6.0 mol
1L
= 3.1250 mol !
1L
6.0 mol
VNaOH = 0.52 L
Step 3. Convert the volume of solution to millilitres.
1000 mL
VNaOH = 0.52 L !
1L
VNaOH = 520 mL
Statement: The volume of a 0.45 mol/l sodium hydroxide solution that contains 125 g of sodium
hydroxide is 520 mL.
Section 8.6 Questions, page 402
1. (a) Concentration is used to describe the quantity of solute per unit volume of solution. The
solubility of a solution is expressed as the mass of solute required to form a saturated solution in
100 g of water at a given temperature.
(b) A concentrated solution contains a relatively large quantity of solute per unit volume of
solution while a saturated solution contains the maximum quantity of solute at a given
temperature and pressure.
2. An acid solution can be diluted to half its original concentration by adding water until the final
volume of the mixture is twice the original volume.
3. Given: volume of silver nitrate, VAgNO = 500.0 mL
3
mass of silver nitrate, mAgNO = 34.0 g
3
Required: amount concentration of silver nitrate solution, cAgNO
Analysis: cAgNO =
3
nAgNO
3
VAgNO
3
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Chapter 8: Water and Solutions
8.6-3
Solution:
Step 1. Convert the given volume into litres.
1L
VAgNO = 500.0 mL !
3
1000 mL
VAgNO = 0.5000 L
3
Step 2. Convert the mass of solute to amount of solute.
1 mol
nAgNO = 34.0 g !
3
169.88 g
nAgNO = 0.200 14 mol
3
Step 3. Substitute the values and solve the equation.
nAgNO
3
cAgNO =
3
VAgNO
3
0.200 14 mol
0.5000 L
mol
cAgNO = 0.400
3
L
Statement: The amount concentration of silver nitrate solution is 0.400 mol/L.
4. Given: concentration of potassium hydroxide, cKOH = 2.0 mol/L
=
volume of potassium hydroxide, VKOH = 1.5 L
Required: mass of potassium hydroxide, mKOH
Analysis: cKOH =
nKOH
VKOH
Solution:
Step 1. Rearrange the formula to calculate amount.
nKOH = cKOHVKOH
2.0 mol
! 1.5 L
1L
nKOH = 3.0 mol
Step 2. Determine the mass of potassium hydroxide.
56.11 g
mKOH = 3.0 mol !
1 mol
mKOH = 170 g
Statement: The mass of potassium hydroxide required to prepare 1.5 L of a 2.0 mol/L solution
is 170 g.
=
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Chapter 8: Water and Solutions
8.6-4
5. Given: concentration of sodium carbonate, cNa CO = 0.45 mol/L
2
3
mass of sodium carbonate, mNa CO = 35.0 g
2
3
Required: volume of sodium carbonate solution, VNa CO
2
Analysis: cNa CO =
2
3
nNa CO
3
VNa CO
3
2
2
3
Solution:
Step 1. Determine the molar mass of sodium carbonate.
M Na CO = 2 M Na + M C + 3M O
2
3
!
!
g $
g
g $
= 2 # 22.99
+
12.01
+
3
16.00
#"
mol &%
mol
mol &%
"
g
mol
Step 2. Determine the amount of sodium carbonate.
1 mol
nNa CO = 35.0 g !
2
3
105.99 g
M Na CO = 105.99
2
3
nNa CO = 0.330 22 mol [2 extra digits carried]
2
3
Step 3. Rearrange the formula to determine volume.
nNa CO
2
3
VNa CO =
2
3
cNa CO
2
=
3
0.33022 mol
0.45 mol
1L
= 0.330 22 mol !
1L
0.45 mol
VNa CO = 0.73 L
2
3
Statement: The volume of a 0.45 mol/L sodium carbonate solution that contains 35.0 g of
sodium carbonate is 730 mL.
6. Given: concentration of sodium oxalate, cNa C O = 0.100 mol/L
2
2
4
mass of sodium oxalate, mNa C O = 33.5 g
2
2
4
Required: volume of sodium oxalate solution, VNa C O
2
Analysis: cNa C O =
2
2
4
nNa C O
4
VNa C O
4
2
2
2
2
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2
4
Chapter 8: Water and Solutions
8.6-5
Solution:
Step 1. Determine the molar mass of sodium oxalate.
M Na C O = 2 M Na + 2 M C + 4 M O
2
2
4
!
!
!
g $
g $
g $
= 2 # 22.99
+
2
12.01
+
4
16.00
#"
#"
mol &%
mol &%
mol &%
"
g
mol
Step 2. Determine the amount of sodium oxalate.
1 mol
nNa C O = 33.5 g !
2 2 4
134.00 g
M Na C O = 134.00
2
2
4
nNa C O = 0.2500 mol [2 extra digits carried]
2
2
4
Step 3. Rearrange the formula to calculate the volume.
nNa C O
2 2 4
VNa C O =
2 2 4
cNa C O
2
=
2
4
0.2500 mol
0.100 mol
1L
= 0.2500 mol !
1L
0.1 mol
VNa C O = 2.50 L
2
2
4
Statement: The volume of a 0.100 mol/l sodium oxalate solution that contains 33.5 g of
sodium oxalate is 2.50 × 103 mL.
7. Given: concentration of copper sulfate, cCuSO = 0.20 mol/L
4
volume of copper sulfate, VCuSO = 450 mL
4
Required: mass of copper(II) sulfate pentahydrate, mCuSO
Analysis: cCuSO =
4
nCuSO
4
VCuSO
4
4 •5H 2 O
Solution:
Step 1. Convert the volume to litres.
1L
VCuSO = 450 mL !
4
1000 mL
VCuSO = 0.450 L
4
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Chapter 8: Water and Solutions
8.6-6
Step 2. Rearrange the formula to calculate amount.
nCuSO = cCuSO VCuSO
4
4
4
0.20 mol
! 0.450 L
1L
= 0.090 mol
=
nCuSO
4
Step 3. Determine the molar mass of copper(II) sulfate pentahydrate.
M CuSO •5H O = M Cu + M S + 9 M O + 10 M H
4
2
= 63.55
!
!
g
g
g $
g $
+ 32.07
+ 9 # 16.00
+ 10 # 1.01
&
mol
mol
mol %
mol &%
"
"
g
mol
Step 4. Determine the mass of copper(II) sulfate pentahydrate.
249.72 g
mCuSO •5H O = 0.090 mol !
4
2
1 mol
mCuSO •5H O = 22 g
M CuSO
4 •5H 2 O
4
= 249.72
2
Statement: The mass of copper(II) sulfate pentahydrate required to prepare 450 mL of a
0.20 mol/L solution is 22 g.
8. Transporting acids as concentrated solutions is more economical because a concentrated acid
solution contains a greater amount (in mol) of acid than a dilute acid. Therefore, a greater
amount of acid can be transported in each shipment.
9. Since the concentration and volume of the solutions is identical, both contain the same amount
(in mole) of acid. The molar mass of sulfuric acid (98.08 g/mol) is greater than the molar mass of
acetic acid (60.06 g/mol). Therefore, the mass of 1 L of 18 mol/L sulfuric acid is greater than the
mass of 1 L of 18 mol/L acetic acid.
10. Given: concentration of cholesterol, cC H O = 6.2 ! 10 "3 mol/L
27
46
volume of blood, Vblood = 4.7 L
Required: mass of cholesterol, mC
Analysis: cC
27 H 46 O
=
nC
27 H 46 O
VC
27 H 46 O
27 H 46 O
Solution:
Step 1. Rearrange the formula to calculate amount.
nC H O = cC H OVC H O
27
46
27
=
nC
27 H 46 O
46
27
46
"3
6.2 ! 10 mol
! 4.7 L
1L
= 2.914 ! 10 "2 mol [2 extra digits carried]
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Chapter 8: Water and Solutions
8.6-7
Step 2. Determine the molar mass of cholesterol.
M C H O = 27 M C + 46 M H + M O
27
46
!
!
g $
g $
g
= 27 # 12.01
+ 46 # 1.01
+ 16.00
&
&
mol %
mol %
mol
"
"
g
mol
Step 3. Determine the mass of cholesterol.
386.73 g
mC H O = 2.914 ! 10 "2 mol !
27 46
1 mol
mC H O = 11 g
MC
27 H 46 O
27
= 386.73
46
Statement: The mass of cholesterol in 4.7 L of blood is 11 g.
11. The narrow neck of the volumetric flask makes it easier for chemists to make a precise
reading of the water level in the flask. The larger width of the beaker makes it more difficult to
make precise volume measurements. Therefore, a volumetric flask is preferred over a beaker
when preparing solutions of precise concentration.
12. (a) It is important to store standard solutions in sealed containers to prevent the evaporation
of solvent and to prevent chemicals in the air from contaminating the solution.
(b) Some of the solvent may evaporate if a sodium chloride solution is stored in an open
container. As this occurs, the overall volume, V, of the solution decreases, resulting in an
n
increase in amount concentration, c, because c = .
V
13. One benefit of having a recognized organization such as IUPAC deciding on naming
conventions is that the common terminology will be used by all chemists. This eliminates any
confusion that may arise from the use of terms that may not be familiar to everyone. One
drawback is that the new terminology may not be as familiar to everyone.
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Chapter 8: Water and Solutions
8.6-8