Section 8.8: Concentrations and Consumer Products
Tutorial 1 Practice, page 408
1. Given: cV/V = 10 %
Vsolution = 40 L
Required: volume of ethanol, Vethanol
Analysis: cV/V =
Vethanol
! 100 %
Vsolution
Solution: Rearrange the equation in the appropriate form, substitute the values, and solve.
c
Vethanol = Vsolution ! V/V
100 %
10 %
= 40 L !
100 %
Vethanol = 4.0 L
Statement: The volume of ethanol in a 40 L fill-up of gasoline is 4.0 L.
2. Given: mNaClO = 125 g
Vsolution = 2.5 L
Required: percentage weight/volume of bleach, cW/V
Analysis: cW/V =
mNaClO
Vsolution
! 100 %
Solution:
Step 1. Convert the volume of solution to millilitres.
1000 mL
Vsolution = 2.5 L !
1L
Vsolution = 2.5 ! 103 mL
Step 2. Substitute the values into the equation and solve.
125 g
cW/V =
! 100 %
2.5 ! 103 mL
cW/V = 5.0 %
Statement: The percentage weight/volume of bleach is 5.0 % W/V.
3. Given: cW/W = 2.5 %
Required: mass of boric acid, msolute
Analysis: cW/W =
msolute
msolution
! 100 %
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8.8-1
Solution: Rearrange the equation in the appropriate form, substitute the values, and solve.
c
msolute = msolution ! W/W
100 %
2.5 %
= 125 g !
100 %
msolute = 3.1 g
Statement: The mass of boric acid present in 125 g of a 2.5 % W/W solution is 3.1 g.
4. Given: cc = 30.0 % W/V
cd = 6.0 % W/V
Vd = 425 mL
Required: volume of the concentrated solution, Vc
Analysis: ccVc = cdVd
Solution: Rearrange the equation in the appropriate form, substitute the values, and solve.
cV
Vc = d d
cc
=
6.0 %W/V ! 425 mL
30.0 %W/V
Vc = 85 mL
Statement: The volume of initial concentration of hydrogen peroxide solution is 85 mL.
Tutorial 2 Practice, page 410
1. Given: cppm = 250 ppm
mshampoo = 500 g
Required: mass of methanal, mmethanal
Analysis: cppm =
msolute
msolution
! 10
6
Solution: Rearrange the equation in the appropriate form, substitute the values, and solve.
cppm ! mshampoo
mmethanal =
106
250 ! 500 g
=
106
mmethanal = 0.13 g
Statement: The mass of methanal in the shampoo is 0.13 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8.8-2
2. (a) Given: msolute = 1.5 mg
Vwater = 1.0 L
Required: fluoride concentration in parts per million, cppm
Analysis: cppm =
msolute
msolution
! 106
Solution:
Step 1. Convert 1.0 L of water to mass.
Since the density of water is 1.0 g/mL, then 1.0 L of water has a mass of 10 kg.
1.0 kg = 1.0 × 106 mg
Step 2. Substitute the values and solve.
1.5 mg
cppm =
! 106
6
1.0 ! 10 mg
cppm = 1.5 ppm
Statement: A concentration of 1.5 mg/L of fluoride in water corresponds to 1.5 ppm.
(b) Given: cppm = 1.5 ppm
msolute = 1.0 g
Required: volume of tap water, Vwater
Analysis: cppm =
msolute
msolution
! 106
Solution:
Step 1. Rearrange the equation in the appropriate form, substitute the values, and solve.
m
msolution = solute ! 106
cppm
=
1.0 g
! 106
1.5 ppm
msolution = 6.7 ! 105 g
Step 2. Calculate the volume of water.
Since the density of water is 1.0 g/mL, the volume of water is 6.7 × 105 mL.
1L
Vwater = 6.7 ! 105 mL !
1000 mL
Vwater = 670 L
Statement: The volume of water containing 1.0 g of fluoride is 670 L.
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8.8-3
3. Given: Vwater = 250 mL
cppb = 12 ppb
Required: mass of antibiotic in sample, msolute
Analysis: cppb =
msolute
msolution
! 109
Solution:
Step 1. Calculate the mass of solution.
Since the density of water is 1.0 g/mL, the mass of water is 250 g.
Step 2. Rearrange the equation in the appropriate form, substitute the values, and solve.
cppb ! msolution
msolute =
109
12 ! 250 g
=
109
msolute = 3.0 ! 10 "6 g
Statement: The mass of antibiotic in the sample is 3.0 × 10–6 g.
Research This: The BPA Controversy, page 410
A. The companies that manufacture BPA will likely have to invest in research and development
to develop safer alternatives to BPA. This will reduce their profitability.
B. Answers may vary. Sample answers are: canned soft drinks, 0.11 µg/L; canned tuna, nonpolycarbonate baby bottles, 0.0024 µg/L (after 2 h high temperature).
C. Based on the current research, there is enough concern about the long-term health risk of BPA
exposure to warrant the “toxic” designation. Statistics Canada reported that 91 % of all people
from 6 to 79 years had BPA in their urine (mean concentration of 1.16 µg/L in their urine).
Studies need to be done to determine the long-term health effects of BPA.
D. Answers may vary. There are a number of non-plastic containers that are suitable
replacements for BPA-containing products. Glass bottles, for example, could replace plastic
bottles.
Section 8.8 Questions, page 411
1. Given: cppm = 3.0 ppm
VH O = 3.4 ! 106 L
2
Required: mass of chlorine, mCl
Analysis: cppm =
msolute
msolution
! 10
6
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8.8-4
Solution:
Step 1. Calculate the mass of the solution.
Assuming the density of the pool water is 1.0 g/mL, the mass of the pool water is 3.4 × 106 g.
Step 2. Rearrange the equation in the appropriate form, substitute the values, and solve.
cppm ! mH O
2
mCl =
6
10
=
3.0 ! 3.4 ! 10
6
106
mCl = 10 g
Statement: The mass of chlorine in the pool is 10 g.
2. Given: mmethanal = 250 g
Required: mass of water, mH O
2
Analysis: cmethanal =
mmethanal
msolution
cmethanal = 37 % W/W or 37 g / 100 g
Solution:
Step 1. Calculate the mass of the solution.
Assuming the density of the pool water is 1.0 g/mL, the mass of the pool water is 3.4 × 106 g.
Step 2. Rearrange the equation in the appropriate form, substitute the values, and solve.
msolution =
=
mmethanal
cmethanal
250 g methanal
37 g methanal
100 g solution
= 250 g methanal !
100 g solution
37 g methanal
msolution = 680 g
Step 3. Calculate the mass of the water.
mH O = 680 g – 250 g
2
mH O = 430 g
2
Statement: The mass of water that must be added is 430 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8.8-5
3. Given: Vsolution = 1.5 L
c2-propanol = 70 % V/V or 70 mL / 100 mL
Required: volume of 2-propanol, V2-propanol
Analysis: c2-propanol =
V2-propanol
Vsolution
Solution:
Step 1. Rearrange the equation to solve for volume of 2-propanol.
V2-propanol = c2-propanol ! Vsolution
Step 2. Substitute the correct values and solve.
V2-propanol = 0.70 ! 1.5 L
V2-propanol = 1.0 L
Statement: A 1.5 L solution of rubbing alcohol contains 1.0 L of 2-propanol.
4. (a) Given: mglucose = 45 g
Vsolution = 250 mL
Required: percentage concentration of the solution, cglucose
Analysis: cglucose =
mglucose
! 100 %
Vsolution
Solution: Substitute the correct values into the equation and solve.
45 g
cglucose =
! 100%
250 mL
cglucose = 18 % W/V
Statement: The percentage concentration of the glucose solution is 18% W/V.
(b) Given: mglucose = 45 g
Vsolution = 250 mL
Required: concentration of the solution, cglucose
Analysis: cglucose =
nglucose
Vglucose
Solution:
Step 1. Convert volume from millilitres to litres.
1L
Vglucose = 250 mL !
1000 mL
Vglucose = 0.25 L
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8.8-6
Step 2. Determine the molar mass of glucose.
M C H O = 6 M C + 12 M H + 6 M O
6
12
6
!
!
!
g $
g $
g $
= 6 # 12.01
+ 12 # 1.01
+ 6 # 16.00
&
&
mol %
mol %
mol &%
"
"
"
g
6
mol
Step 3. Determine the amount of glucose.
1 mol
nglucose = 45 g !
180.18 g
MC H
12 O6
= 180.18
nglucose = 0.2498 mol [2 extra digits carried]
Step 4. Determine the concentration of glucose.
0.2498 mol
cglucose =
0.25 L
mol
cglucose = 1.0
L
Statement: The amount concentration of the glucose solution is 1.0 mol/L.
5. (a) Since 1 ppm is equal to 1 g of solute in every 1 × 106 L, 30.08 mg/L is equivalent to
30.08 ppm.
(b) Given: cibuprofen = 30.08 ppm
mibuprofen = 1.00 g
Required: volume of water that contains 1.00 g ibuprofen, Vwater
Analysis: cibuprofen =
mibuprofen
mwater
! 106
Solution:
Step 1. Rearrange the equation to solve for the mass of the water.
106
mwater = mibuprofen !
cibuprofen
Step 2. Substitute the correct values and solve.
106
mwater = 1.00 g !
30.08 ppm
4
mwater = 3.32 ! 10 g
Step 3. Determine the mass of the water.
Since the density of water is 1.0 g/mL, 3.32 × 104 g of water corresponds to 3.32 × 104 mL
of water.
Statement: The volume of water that contains 1.00 g of ibuprofen is 3.32 × 104 mL, or 33.2 L.
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8.8-7
6. Given: mTriclosan = 38 ng
VTriclosan = 1.5 L
Required: concentration of the solution, cppm
Analysis: cppb =
mTriclosan
msolution
! 109
Solution:
Step 1. Determine the mass of the water.
Since the density of water is 1.0 g/mL, the mass of the water is 1500 g, or 1.5 × 103 g.
Step 2. Convert the mass from nanograms to grams.
1g
mTriclosan = 38 ng ! 9
10 ng
mTriclosan = 3.8 ! 10 "8 g
Step 3. Determine the concentration.
3.8 ! 10 "8 g
cppb =
! 109
3
1.5 ! 10 g
cppb = 2.5 ! 10 –2 ppb
Statement: The concentration of Triclosan in the water is 2.5 × 10–2 ppb.
7. A 1.0 mol/L solution of sodium chloride is more concentrated than a 1 % W/W solution of
sodium chloride. One litre of 1.0 mol/L solution of sodium chloride contains 1 mol or 58.45 g of
sodium chloride. A 1 % W/W sodium chloride solution is a dilute solution. Since the density of
this solution is similar to the density of water, which is 1.0 g/mL, one litre of this solution has a
mass of about 1000 g. One percent of this mass is 10 g of sodium chloride. Therefore, the
1.0 mol/L solution of sodium chloride contains more sodium chloride and is more concentrated.
8. It is important to standardize the way in which concentrations of industrial chemicals are
expressed to ensure that there is no ambiguity in the amount of the chemical being used or sold.
9. (a) The mass of sodium hydrogen carbonate in the bottle is 4.2 g.
1 mol
(b) nNaHCO = 4.2 g !
3
84.01 g
nNaHCO = 0.050 mol
3
The amount of sodium hydrogen carbonate in the bottle is 0.050 mol.
mNaHCO
3
! 100 %
(c) cW/V =
Vsolution
4.2 g
! 100 %
50 mL
cW/V = 8.4 %
The concentration of sodium hydrogen carbonate is 8.4 % W/V.
cW/V =
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8.8-8
(d) Since the density of water is 1.0 g/mL, the mass of the water is 50 g.
m
cppm = solute ! 106
msolution
4.2 g
! 106
50 g
= 84 000 ppm
cppm =
cppm
The concentration of sodium hydrogen carbonate is 84 000 ppm.
nNaHCO
3
(e) cNaHCO =
3
VNaHCO
3
=
cNaHCO
3
0.050 mol
"2
5.0 ! 10 L
= 1.0 mol/L
The amount concentration of sodium hydrogen carbonate is 1.0 mol/L.
10. (a) Given: cppt = 12 ppt
gold
8
Vocean = 3.54 ! 10 km 3
d ocean = 1035
kg
3
m
Required: mass of water in the Atlantic Ocean, mocean
Analysis: d ocean =
mocean
Vocean
; d=
m
V
Solution:
Step 1. Rearrange the density equation for the ocean to solve for the mass.
mocean = d ocean ! Vocean
Step 2. Substitute the correct values and solve for the mass, while converting volume.
! 109 m 3 $
!
kg $
8
3
mocean = # 1035
&
& 3.54 ' 10 km ##
3
"
m3 %
" km &%
(
)
mocean = 3.6639 ' 10 20 kg (2 extra digits carried)
Statement: The average mass of water in the Atlantic Ocean is 3.66 × 1020 kg.
20
(b) Given: mocean = 3.6639 ! 10 kg
cppt
gold
= 12 ppt
Required: mass of gold in the Atlantic Ocean, mgold
Analysis: cppt =
mgold
mocean
12
! 10
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8.8-9
Solution:
Step 1. Determine the mass of water.
Since the density of water is 1.0 g/mL, the mass of the water is 1500 g, or 1.5 × 103 g.
Step 2. Rearrange the equation to solve for the mass of gold.
cppt ! mocean
gold
mgold =
1012
Step 3. Substitute the correct values and solve.
12 ! 3.66 ! 10 20 kg
mgold =
1012
mgold = 4.4 ! 109 kg
Statement: The mass of gold in the Atlantic Ocean is 4.4 × 109 kg.
11. (a) Answers may vary. Sample answers: Vinegar is a 5 % V/V solution of ethanoic (acetic
acid in water). Brass is a 85 % W/W copper and 15 % W/W zinc. Antifreeze is 60 % V/V
solution of ethylene glycol in water.
(b) Answers may vary. Sample answer: For vinegar, since both acetic acid and water are liquids,
it is convenient to express the percentage concentration in terms of liquid volumes. Since both
substances are solids in brass, it is convenient to express the percentage concentration in terms of
solid masses. Since both substances are liquids in antifreeze, it is convenient to express the
percentage concentration in terms of liquid volumes.
Copyright © 2011 Nelson Education Ltd.
Chapter 8: Water and Solutions
8.8-10