CHAPTER 5: ANSWERS TO SELECTED PROBLEMS
SAMPLE PROBLEMS (“Try it yourself”)
5.1
Potassium iodide is the solute and alcohol is the solvent.
5.2
H
H
C
N
H
H
δ–
δ+
H
O
H
H
5.3
The oxygen atom in THF is negatively charged, because it has a higher electronegativity
than the two neighboring carbon atoms. As a result, the oxygen atom can act as a hydrogen bond
acceptor. The hydrogen atoms in water are attracted to the oxygen atom in THF, and this
attraction allows THF to mix with water.
5.4
Calcium nitrate dissociates into Ca2+ and NO3– ions. Each of these ions is solvated by
water molecules. The ions are independent of one another and are free to move about the
solution.
5.5
The best choices are Na2CO3 and K2CO3. Most other compounds that contain CO32– do
not dissolve in water.
5.6
90 grams is a significant amount of solute and is easily visible. Therefore, NaHCO3 is
described as a soluble compound.
5.7
Hydrogen sulfide is a gas at room temperature, because its boiling point is far below
25ºC. Gases dissolve best in cold water, so you should cool the water.
5.8
At 75ºC, carbon disulfide is a gas, because its boiling point is 46ºC. Therefore, at 75ºC
the solubility of carbon disulfide increases when you increase the pressure.
5.9
H
H
hydrophilic region
H
H
H
O
H
H
H
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
hydrophobic region
5.10 1,6-diaminohexane has the higher solubility in water, because it has more atoms that can
participate in hydrogen bonds (four hydrogen atoms and two nitrogen atoms).
5.11
4.2% (v/v) (The calculator answer is 4.16666667%.) This is a volume percentage.
5.12
There are 30 g of urea in the bottle.
5.13
2600 mg (2.6 mg). The calculator answer is 2552 mg.
5.14
14.01 g (because 6.022 x 1023 nitrogen atoms is one mole of nitrogen)
5.15
132.144 g/mol
5.16
3.91 g (the calculator answer is 3.9061 g)
5.17
0.442 moles (the calculator answer is 0.44194212 g)
5.18 1.2 M (the calculator answer is 1.21131389 M). The molar mass of glycine is 75.07
g/mol, so we have 0.09084854 moles of glycine in the solution.
5.19 4.2 g (the calculator answer is 4.24775 g). You need 0.025 moles of AgNO3 to prepare
the solution, and the molar mass of AgNO3 is 169.91 g/mol.
5.20 Osmosis does occur, with the water flowing from solution C into solution D. Note that
the total concentration of solutes in solution C is 0.009 M, which is less than the 0.011 M in
solution D.
5.21 The total molarity of solute particles is 0.28 M. The Na2CO3 dissociates into two Na+
ions and one CO32– ion, so the concentration of Na+ in the solution is 2 x 0.07 = 0.14 M, and the
concentration of CO32– in the solution is 0.07 M. The total concentration is 0.14 M Na+ + 0.07
M CO32– + 0.07 M glucose = 0.28 M.
5.22 a) Glucose flows from solution A into solution B, and sucrose flows from solution B into
solution A.
b) Osmosis does not occur, because the total solute molarity is 0.15 M on both sides.
5.23
This question should say “how many grams is this?”
0.166 Eq of NH4+ weighs 2.99 g (The calculator answer is 2.994972 g).
5.24 0.806 Eq/L (the calculator answer is 0.80572963 Eq/L). You have 0.06714414 moles of
Fe3+, which equals 0.20143241 Eq of Fe3+.
5.25 2.5 g of Ca2+ (the calculator answer is 2.505 g). You have a total of 125 mEq of Ca2+,
which equals 0.125 Eq.
5.26
0.8% (v/v). The calculator answer is 0.83333333%.
5.27
You must add 300 mL of water (because your final volume must be 400 mL).
5.28 You must use 30 mL (0.03 L) of the original solution, and you must add 970 mL (0.97 L)
of water to it.
SECTION PROBLEMS
Section 5.1
5.1
NaOH is the solute and water is the solvent.
5.2
You produce a suspension, because flour does not dissolve in water.
5.3
Compounds b and c dissolve in water, because they contain oxygen or nitrogen atoms
that can participate in hydrogen bonds.
5.4
There are two possible types of hydrogen bonds, as shown below.
H
H
H
N
H
δ+
H
N
O
N2H4 is the donor and
water is the acceptor.
δ–
H
H
H
N
N
δ–
H
H
Water is the donor and
N2H4 is the acceptor.
H δ+
O
5.5
H
The two key elements are oxygen and nitrogen.
Section 5.2
5.6
NaNO2 and CuSO4 are ionic compounds, so they are electrolytes.
5.7
MgCl2 dissociates into Mg2+ ions and Cl– ions. These ions are solvated by water
molecules.
5.8
a) Fe3+ and Cl– ions
b) Na+ and SO42– ions
5.9
Both ions are surrounded by water molecules. The oxygen atoms of water face the
potassium ion, and the hydrogen atoms of water face the fluoride ion, as shown below.
H
H
O
H
K+
O
H
H
O
H
H
O
H
H
O
F–
H
O
H
H
H
H
O
H
O
H
5.10 The best choices are NaI and KI. Both compounds are soluble in water, and sodium and
potassium are non-toxic.
Section 5.3
5.11 Only a very small amount of vitamin A can dissolve in a reasonable amount of water (a
liter or a deciliter). The mass of vitamin A that dissolves is so small that it is not evident to an
observer.
5.12 This solubility is very low, so BaSO4 is classified as an insoluble compound. 0.0025
grams is barely detectible to the eye, and would not be noticed.
5.13 You have an unsaturated solution. The maximum mass of aspirin that can dissolve in a
liter of water is 3 grams, so if we dissolve 2.5 grams, we can add still more aspirin to our
solution.
5.14 The solubility of aspirin increases as you increase the temperature. At 35ºC, we can
dissolve more than 3 g of aspirin in a liter of water, so a solution that contains 3 g per liter is
unsaturated.
5.15
A reasonable estimate is between 50ºC and 55ºC.
5.16
A reasonable estimate is around 0.02 g/L.
5.17
a) CaSO4 becomes more soluble as the temperature increases.
b) CO2 becomes less soluble as the temperature increases.
Section 5.4
5.18
H
O
H
H
H
H
H
H
C
N
C
C
C
C
C
H
H
H
H
H
hydrophilic
region
H
hydrophobic region
5.19 a) The first compound has the higher solubility in water. Both compounds contain the
same hydrophilic functional group (the nitrogen atom and the two attached hydrogen atoms), but
the first molecule has a smaller hydrocarbon chain.
b) The second compound has the higher solubility in water. Both compounds have the
same carbon skeleton, but the second molecule contains two hydrophilic functional groups (the
two oxygen atoms), while the first contains only one.
Section 5.5
6 g NH 4 Cl
100 mL solution
5 µg arsenic
d)
1 L solution
2.5 mL H 2SO 4
100 mL solution
90 mg glucose
e)
1 dL solution
10 mg benzene
1 L solution
130 ng ferritin
f)
1 mL solution
5.20
a)
b)
c)
€
5.21
1.4% (w/v)
€
€
€
5.22
24% (v/v)
€
€
5.23
3.7 g of CaCl2 (the calculator answer is 3.75 g)
5.24
1.2 mL of ethanol
5.25
20 mg/dL
5.26
2.5 mg
5.27
20 ppm (Remember that ppm is the same as µg/mL.)
5.28
0.083 ppb (Remember that ppb is the same as ng/mL.)
Section 5.6
5.29
35.45 grams of chlorine. (Both boxes contain one mole.)
5.30 The sample of carbon contains more atoms. From the periodic table, we see that carbon
atoms are lighter than nitrogen atoms. Therefore, we need more carbon atoms than we do
nitrogen atoms in order to have a 12.01 gram sample.
5.31
342.14 amu
5.32
342.14 grams
5.33 0.0622 moles (the calculator answer is 0.06224398 moles). The molar mass of CaCO3 is
100.09 g/mol.
5.34 6.89 grams (the calculator answer is 6.89169 g). The molar mass of NaC3H5O3 is 112.06
g/mol.
Section 5.7
5.35
1.40 M. (You can also write this as 1.40 mol/L.)
5.36 0.0207 M (the calculator answer is 0.02073606 M). The molar mass of MgSO4 is 120.37
g/mol, so you have 0.05184016 moles of MgSO4 in the solution.
5.37
2.5 g of KI (the calculator answer is 2.49 g). The molar mass of KI is 166 g/mol.
5.38 You would need 12 g of MgSO4 (the calculator answer is 12.037 g). You need 0.1 moles
of MgSO4, and the molar mass of MgSO4 is 120.37 g/mol.
Section 5.8
5.39 Chocolate contains a variety of compounds that evaporate easily. Once these compounds
evaporate, they mix with the air in the room until they are evenly distributed throughout the
room. This mixing is an example of diffusion.
5.40 Osmosis is the flow of water through the membrane. Solution A contains a lower
concentration of sucrose than solution B, so it contains a higher concentration of water. In
osmosis, water moves from its highest to its lowest concentration, so water moves from solution
A to solution B.
5.41
a) hypotonic
b) isotonic
c) hypertonic
5.42 a) In the 0.14 M lactose solution, the red blood cell will hemolyze (it will swell and
burst).
b) In the 0.28 M lactose solution, the red blood cell will not change.
c) In the 0.42 M lactose solution, the red blood cell will crenate (it will shrivel).
5.43 NaCl dissociates into Na+ and Cl– ions when it dissolves in water, so the solution actually
contains 0.2 M Na+ and 0.2 M Cl–. The total concentration of solutes in the solution is 0.4 M.
Therefore, the solution is hypertonic, and the cell will crenate.
5.44 The total molarity is 0.42 M. MgCl2 dissociates into one Mg2+ ion and two Cl– ions.
Therefore, when you dissolve 0.14 moles of MgCl2 in water, you form 0.14 moles of Mg2+ ions
and 0.28 moles of Cl– ions (2 x 0.14 = 0.28). Adding these gives the total molarity.
5.45
M.
The solution is hypertonic, because the total concentration of solutions is larger than 0.28
5.46 The molarity of glycine is 0.18 M. An isotonic solution has a total solute concentration
of 0.28 M, so 0.28 M – 0.10 M = 0.18 M.
5.47 Osmosis is the movement of water (the solvent) through a membrane, while dialysis is
the movement of a solute through a membrane.
5.48 The sucrose will move from solution B to solution A, because all substances move from
their highest concentration to their lowest concentration.
Section 5.9
5.49
0.1 equivalents
5.50
a) 0.268 Eq (the calculator answer is 0.26850584 Eq).
b) 268 mEq
5.51
a) 0.127 Eq
c) 4040 mg
5.52
a) 0.0177 Eq/L (the calculator answer is 0.01774039 Eq/L)
b) 17.7 mEq/L
Note that sulfate ion has the formula SO42–, so 96.06 g = 1 mole = 2 Eq.
5.53
The concentration of Ca2+ is 500 mEq/L and the concentration of Cl– is 500 mEq/L.
b) 4.04 g (the calculator answer is 4.035425 g)
Section 5.10
5.54
You need 80 mL of the 2.5% solution, and you must add 120 mL of water.
5.55 The concentration of lactic acid is 6.7 ppm after adding the water. (The calculator
answer is 6.66666667 ppm.)
5.56 The volume of the solution after adding water is 150 mL, so you must add 50 mL of
water.
CUMULATIVE PROBLEMS (Odd-numbered problems only)
5.57
Choice c is the best description.
5.59
Milk of magnesia is a suspension.
5.61 In both cases, the mixture becomes homogeneous. However, when the ice melts, you are
left with a single substance (water). By contrast, when the sugar dissolves, you have a
homogeneous mixture of two substances (sugar and water).
5.63
Choice c is the best explanation.
5.65 There are three possible types of hydrogen bonds that can form between HNO2 and
water.
H
H
δ+
δ–
O
H O
(HNO2 is the donor and
water is the acceptor.)
δ–
H
O
N
O
N
O
H
H
H
δ+
H
(Water is the donor and
HNO2 is the acceptor.)
O
δ–
O
N
H
O
O
(Water is the donor and
HNO2 is the acceptor.)
δ+
5.67 CaCl2 and KC2H3O2 are electrolytes. (They contain a metallic element, so they are ionic
compounds.)
5.69
Choice c is the best explanation.
5.71
a) K+ and S2– ions
b) Fe2+ and SO42– ions
c) NH4+ and CO32– ions
5.73 A substance dissolves when it mixes with the solvent to form a homogeneous mixture. A
substance dissociates if it breaks apart into ions when it dissolves.
5.75
H
O
H
H
The calcium ion is surrounded by
water molecules, with the negatively
charged oxygen atoms facing the ion.
H
Ca2+
O
H
O
H
O
H
5.77
H
The best choices are Na2MoO4 or K2MoO4.
5.79 Methanol contains a positively charged hydrogen atom. When chloride ions dissolve in
methanol, the methanol molecules surround the ions, with the positively charged hydrogen atoms
facing the ions, as shown below.
H
H C H
O
H
H
H C
O H
Cl
–
H
H O
C H
H
H
H
O
H C H
H
5.81 CH2Cl2 is not polar enough to solvate ions, so it is a very poor solvent for ionic
compounds. It also cannot form hydrogen bonds, so it cannot mix with water or other substances
that form many hydrogen bonds. Therefore, CH2Cl2 probably could not serve as the primary
biological solvent.
5.83 To compare the solubility of niacin with the solubilities that are listed in Table 5.2, we
must calculate the solubility of niacin in one liter of water. In this unit, the solubility is 17 grams
per liter (g/L), which is high enough to classify niacin as a soluble compound.
5.85 a) HCl is a gas between 0ºC and 60ºC. Notice that the solubility decreases as you
increase the temperature. This is typical of a gas; solids usually become more soluble as you
increase the temperature.
b) Since HCl is a gas, its solubility should increase as the pressure increases.
c) The solubility is roughly 700 g/L at 25ºC.
d) The highest temperature you could use is roughly 50ºC. Above this temperature, the
solubility of HCl is less than 600 g/L.
5.87
H
H
N
C
C
C
H
hydrophilic regions
niacin (one of the B vitamins)
C
C
C
N
H
O
H
H
5.89 a) The second compound has the higher solubility, because it has a shorter hydrocarbon
chain.
b) The second compound has the higher solubility, because it has two oxygen atoms
(each of which can participate in hydrogen bonds).
c) The second compound has the higher solubility, because it has a hydrogen bonding
group, while the first molecule does not.
5.91 Magnesium carbonate does not dissolve in water, so it cannot be used to prepare a
solution that contains Mg2+ ions. Magnesium chloride is water-soluble.
€
5.95
0.15 mL acetic acid
50 mg Br b)
100 mL solution
1 L solution
6.5 g MgSO 4
20 µg Pb 2+
d)
e)
100 mL solution
1 L solution
€
€
9.2% (w/v). (The calculator answer is 9.24%.)
€
5.97
€
72.8% (v/v). (The calculator
answer is 72.83950617%.)
5.93
a)
c)
3 mg fructose
1 dL solution
5.99
You must use 3.1 grams of vitamin C. (The calculator answer is 3.125 g.)
5.101 94 mL of ethylene glycol was used. (The calculator answer is 93.75 mL.)
5.103 Blood plasma contains 70 mg of Mg2+.
5.105 The bicarbonate concentration is 160 mg/dL. (The calculator answer is 157.1428571
mg/dL.)
5.107 1040 mg, or 1.04 g.
5.109 You are consuming 2 mg of fluoride.
5.111 a) 0.051% (w/v). (The calculator answer is 0.05084746%.)
b) 51 mg/dL. (The calculator answer is 50.84745763 mg/dL.)
c) 510 ppm. (The calculator answer is 508.4745763 ppm.)
5.113 This question should refer back to Problem 5.111, not to Problem 5.107. The percentage
is a poor choice, because it gives a very small number. (The best choice is mg/dL.)
5.115 a) 64.06 amu
b) 64.06 g
5.117 132.144 amu
5.119 78.004 amu. Aluminum hydroxide has the formula Al(OH)3.
5.121 The sample weighs 30.97 g.
5.123 You have 0.0762 moles of leucine. (The calculator answer is 0.07623462 moles.)
5.125 You have 22.9 g of saccharin. (The calculator answer is 22.8975 g.)
5.127 a) 0.038 mol
b) 2.2 g (The calculator answer is 2.22072 g.)
c) 2200 mg (The calculator answer is 2220.72 mg.)
5.129 Chemical formulas give the numbers of atoms, not the masses. Water contains two
hydrogen atoms for every one oxygen atom, but it does not contain two grams of hydrogen for
every gram of oxygen, because hydrogen atoms are much lighter than oxygen atoms.
5.131 a) The sample in box B is heavier.
b) Both samples contain the same number of molecules (6.022 x 1023).
c) The sample in box B contains more atoms, because each CO2 molecule contains three
atoms, while each CO molecule only contains two atoms.
5.133 0.117 M. (The calculator answer is 0.1170234 M.)
5.135 0.4 g of NaOH. (The calculator answer is 0.39998 g.)
5.137 2.0 liters. (The calculator answer is 1.9824089 L.)
5.139 1.73% (w/v). (The calculator answer is 1.731362%.)
5.141 0.0933 M.
5.143 0.26 M. NaCl dissociates into Na+ and Cl–, so the solution contains 0.1 M Na+ and 0.1 M
Cl–. CaCl2 dissociates into Ca2+ and two Cl–, so the solution contains 0.02 M Ca2+ and an
additional 0.04 M Cl–. Adding all of these numbers gives the total molarity.
5.145 0.12 M. The KCl dissociates into K+ and Cl–, so the KCl gives us 0.08 M K+ and 0.08 M
Cl–, for a total of 0.16 M ions. The total molarity of an isotonic solution is around 0.28 M, so the
molarity of glucose must be 0.28 – 0.16 = 0.12 M.
5.147 6.57 g of glucose. (The calculator answer is 6.56947131 g.) The molarity of KCl is
0.06706908 M. KCl dissociates completely, so the solution contains 0.06706908 moles/L of K+
and of Cl–. Since the isotonic concentration is 0.28 M, the concentration of glucose must be
0.14586184 M.
5.149 When these solutes dissolve, they dissociate completely. The concentration of each ion is
0.15 M, so the total concentration of ions in the solution is around 0.3 M, roughly the isotonic
value.
5.151 Note that the total concentration of solutes is 0.25 M in solution A, but 0.2 M in solution
B.
a) Osmosis occurs from right to left, because solution B has a higher concentration of
water than solution A.
b) Na+ dialyzes from left to right, because solution A has a higher concentration of Na+
than solution B.
c) Glucose dialyzes from right to left, because solution B has a higher concentration of
glucose than solution A.
5.153 a) Osmosis does not occur, because both solutions have the same overall concentration of
solutes.
b) Glucose dialyzes from left to right. (Starch cannot pass through the membrane, so it
does not dialyze.)
c) After a few minutes, osmosis occurs from left to right. As glucose passes through the
membrane, the concentration of solutes increases in solution F. Therefore, the concentration of
water becomes lower in solution F (and it becomes higher in solution E). Since there is now a
higher concentration of water in solution E than in solution F, water flows from E to F.
5.155 a) approximately isotonic
b) hypotonic
5.157 a) The cell hemolyzes (it swells and bursts).
c) hypertonic
b) The cell remains unchanged.
c) The cell crenates (it shrivels up.)
5.159 You have 0.4 Eq of CO32–, which equals 400 mEq.
5.161 a) 0.070 moles
b) 0.0080 moles
5.163 The solution contains 0.17 Eq of Ca2+ (the calculator answer is 0.1746507 Eq). This is
equal to 170 mEq (the calculator answer is 174.6507 mEq).
5.165 a) 0.08 Eq
b) 0.04 mol c) 40 mmol d) 2.6 g
e) 2600 mg
(The calculator answers to d and e are 2.6164 g and 2616.4 mg, respectively.)
5.167 0.55 grams of potassium (the calculator answer is 0.5474 g)
5.169 a) 0.0125 M
b) 25 mEq/L
c) 50 mg/dL
d) 500 ppm
5.171 a) Cell membranes only allow certain solutes to pass through. In this case, glucose-6phosphate cannot pass through the membrane, so none of it escapes from the cell into the plasma.
b) Water does not enter the cells because the total concentration of solutes inside and
outside the cell is equal. Glucose-6-phosphate is just one of many solutes inside the cell.
5.173 The final volume of the solution is 291 mL, so you must add 191 mL of water. (The
calculator answer for the final volume is 290.6976744 mL.)
5.175 You must use 34.4 mL of the 5% solution.
5.177 The concentration of iron is 10 ppm.
5.179 The concentration of lactic acid is 5.9 ppm (the calculator answer is 5.88235294 ppm).
5.181 The concentration of citric acid is 0.36% (w/v). (The calculator answer is 0.35511364%).
5.183 The final volume of the solution is 357 mL, so you must add 307 mL of water. (The
calculator answer for the final volume is 357.1428571 mL.)
5.185 The molarity is 0.10 M.