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CHAPTER 14
SOLUTIONS
SOLUTIONS TO REVIEW QUESTIONS
1.
O
H
H
H
H
H
H
H
O
O
Cl–
H
O
H
O
O
H
O
H
H
H
Na+
H
H
H
O
These diagrams are intended to illustrate the orientation of the water molecules about the
ions, not the number of water molecules.
2.
From Table 14.2, approximately 4.5 g of NaF would be soluble in 100 g of water at 50°C.
3.
From Figure 14.4, solubilities in water at 25°C are:
(a) KCl
35 g>100 g H2O
9 g>100 g H2O
(b) KClO3
39 g>100 g H2O
(c) KNO3
4.
Potassium fluoride has a relatively high solubility when compared to lithium or sodium
fluoride. For lithium and sodium halides, the order of solubility (in order of increasing
solubilities) is:
F - Cl- Br - I For potassium halides, the order of increasing solubilities is:
Cl- Br - F - I KClO3 at 60°C, 25 g
HCl at 20°C, 72 g
5.
(a)
(b)
6.
KNO3
(c)
(d)
- 180 -
Li 2SO4 at 80°C, 30 g
KNO3 at 0°C, 14 g
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7.
A one molal solution in camphor will show a greater freezing point depression than a
2 molal solution in benzene.
¢t f = a
1 mol solute 40°C kg camphor
ba
b = 40°C (freezing point depression)
kg camphor
mol solute
2 mol solute 5.1°C kg benzene
¢tf = a
ba
b = 10.2°C (freezing point depression)
kg benzene
mol solute
8.
Cube
1 cm
0.01 cm
Volume
1 cm3
1 * 10-6 cm3
Number> 1 cm cube
1
Area of face
1 cm2
Total surface area
6 cm2
106 311 cm32>11 * 10-6 cm32 = 106 cubes4
1 * 10-4 cm2
6 * 102 cm2
11 * 106 cubes216 faces>cube211 * 10-4 cm2>face2 = 6 * 102 cm2
9.
63 g NH 4Cl
42 g NH 4Cl
From Figure 14.4, the solubility of NH 4Cl in water is
=
150 g H 2O
100 g H 2O
approximately 42 g>100 g H 2O at 30°C, 46 g>100 g H 2O at 40°C. Therefore, the solution
of 63 g>150 g of water would be saturated at 10°C, 20°C, and 30°C. The solution would
be unsaturated at 40°C and 50°C.
10.
The dissolving process involves solvent molecules attaching to the solute ions or
molecules. This rate decreases as more of the solvent molecules are already attached to
solute molecules. As the solution becomes more saturated, the number of unused solvent
molecules decreases. Also, the rate of recrystallization increases as the concentration of
dissolved solute increases.
11.
A supersaturated solution of NaC2H 3O2 may be prepared in the following sequence:
(a)
(b)
(c)
(d)
Determine the mass of NaC2H 3O2 necessary to saturate a specific amount of water
at room temperature.
Place a bit more NaC2H 3O2 in the water than the amount needed to saturate the
solution.
Heat the solution until all the solid dissolves.
Cover the container and allow it to cool undisturbed. The cooled solution, which
should contain no solid NaC2H 3O2 , is supersaturated.
To test for supersaturation, add one small crystal of NaC2H 3O2 to the solution. Immediate
crystallization is an indication that the solution was supersaturated.
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12.
Because the concentration of water is greater in the thistle tube, the water will flow
through the membrane from the thistle tube to the urea solution in the beaker. The
solution level in the thistle tube will fall.
13.
A true solution is one in which the size of the particles of solute are between 0.1 - 1nm.
True solutions are homogeneous and the ratio of solute to solvent can be varied. They can
be colored or colorless but are transparent. The solute remains distributed evenly in the
solution, it will not settle out.
14.
The two components of a solution are the solute and the solvent. The solute is dissolved
into the solvent or is the least abundant component. The solvent is the dissolving agent or
the most abundant component.
15.
It is not always apparent which component in a solution is the solute. For example, in a
solution composed of equal volumes of two liquids, the designation of solute and solvent
would be simply a matter of preference on the part of the person making the designation.
16.
The ions or molecules of a dissolved solute do not settle out because the individual
particles are so small that the force of molecular collisions is large compared to the force
of gravity.
17.
Yes. It is possible to have one solid dissolved in another solid. Metal alloys are of this
type. Atoms of one metal are dissolved among atoms of another metal.
18.
Orange. The three reference solutions are KCl, KMnO4 , and K 2Cr2O7 . They all contain
K + ions in solution. The different colors must result from the different anions dissolved in
the solutions: MnO4 - (purple) and Cr2 O7 2- (orange). Therefore, it is predictable that the
Cr2O7 2- ion present in an aqueous solution of Na 2Cr2O7 will impart an orange color to
the solution.
19.
Hexane and benzene are both nonpolar molecules. There are no strong intermolecular
forces between molecules of either substance or with each other, so they are miscible.
Sodium chloride consists of ions strongly attracted to each other by electrical attractions.
The hexane molecules, being nonpolar, have no strong forces to pull the ions apart, so
sodium chloride is insoluble in hexane.
20.
Coca Cola has two main characteristics, taste and fizz (carbonation). The carbonation is
due to a dissolved gas, carbon dioxide. Since dissolved gases become less soluble as
temperature increases, warm Coca Cola would be flat, with little to no carbonation. It is,
therefore, unappealing to most people.
21.
Air is considered to be a solution because it is a homogeneous mixture of several gaseous
substances and does not have a fixed composition.
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22.
A teaspoon of sugar would definitely dissolve more rapidly in 200 mL of hot coffee than
in 200 mL of iced tea. The much greater thermal agitation of the hot coffee will help
break the sugar molecules away from the undissolved solid and disperse them throughout
the solution. Other solutes in coffee and tea would have no significant effect. The
temperature difference is the critical factor.
23.
The solubility of gases in liquids is greatly affected by the pressure of a gas above the
liquid. The greater the pressure, the more soluble the gas. There is very little effect of
pressure regarding the dissolution of solids in liquids.
24.
For a given mass of solute, the smaller the particles, the faster the dissolution of the
solute. This is due to the smaller particles having a greater surface area exposed to the
dissolving action of the solvent.
25.
In a saturated solution, the net rate of dissolution is zero. There is no further increase in
the amount of dissolved solute, even though undissolved solute is continuously
dissolving, because dissolved solute is continuously coming out of solution, crystallizing
at a rate equal to the rate of dissolving.
26.
When crystals of AgNO3 and NaCl are mixed, the contact between the individual ions is
not intimate enough for the double displacement reaction to occur. When solutions of the
two chemicals are mixed, the ions are free to move and come into intimate contact with
each other, allowing the reaction to occur easily. The AgCl formed is insoluble.
27.
A nonvolatile solute (such as salt) lowers the freezing point of water. Adding salt to icy
roads in winter melts the ice because the salt lowers the freezing point of water.
28.
A 16 molar solution of nitric acid is a solution that contains 16 moles HNO3 per liter of
solution.
29.
The two solutions contain the same number of chloride ions. One liter of 1 M NaCl
contains 1 mole of NaCl, therefore 1 mole of chloride ions. 0.5 liter of 1 M MgCl 2
contains 0.5 mol of MgCl 2 and 1 mole of chloride ions.
10.5 L2 ¢
30.
1 mol MgCl 2
2 mol Cl≤¢
≤ = 1 mol ClL
1 mol MgCl 2
The champagne would spray out of the bottle all over the place. The rise in temperature
and the increase in kinetic energy of the molecules by shaking both act to decrease the
solubility of gas within the liquid. The pressure inside the bottle would be great. As the
cork is popped, much of the gas would escape from the liquid very rapidly, causing
the champagne to spray.
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31.
The number of grams of NaCl in 750 mL of 5.0 molar solution is
10.75 L2a
5.0 mol NaCl 58.44 g
ba
b = 2.2 * 102 g NaCl
L
1 mol
Dissolve the 220 g of NaCl in a minimum amount of water, then dilute the resulting
solution to a final volume of 750 mL (0.75 L).
32.
A semipermeable membrane will allow water molecules to pass through in both
directions. If it has pure water on one side and 10% sugar solutions on the other side of
the membrane, there is a higher concentration of water molecules on the pure water side.
Therefore, there are more water molecule impacts per second on the pure water side of
the membrane. The net result is more water molecules pass from the pure water to the
sugar solution. Osmotic pressure effect.
33.
The urea solution will have the greater osmotic pressure because it has 1.67 mol
solute> kg H 2O, while the glucose solution has only 0.83 mol solute> kg H 2O.
34.
A lettuce leaf immersed in salad dressing containing salt and vinegar will become limp
and wilted as a result of osmosis. As the water inside the leaf flows into the dressing
where the solute concentration is higher the leaf becomes limp from fluid loss. In water,
osmosis proceeds in the opposite direction flowing into the lettuce leaf maintaining a high
fluid content and crisp leaf.
35.
The concentration of solutes (such as salts) is higher in seawater than in body fluids. The
survivors who drank seawater suffered further dehydration from the transfer of water by
osmosis from body tissues to the intestinal tract.
36.
Ranking of the specified bases in descending order of the volume of each required to
react with 1 liter of 1 M HCl. The volume of each required to yield 1 mole of OH - ion is
shown.
(a) 1 M NaOH
1 liter
(b) 0.6 M Ba(OH)2
0.83 liter
(c) 2 M KOH
0.50 liter
(d) 1.5 M Ca(OH)2
0.33 liter
37.
The boiling point of a liquid or solution is the temperature at which the vapor pressure
of the liquid equals the pressure of the atmosphere. Since a solution containing a
nonvolatile solute has a lower vapor pressure than the pure solvent, the boiling point of
the solution must be at a higher temperature than for the pure solvent. At the higher
boiling temperature the vapor pressure of the solution equals the atmospheric pressure.
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38.
The freezing point is the temperature at which a liquid changes to a solid. The vapor
pressure of a solution is lower than that of a pure solvent. Therefore, the vapor pressure
curve of the solution intersects the vapor pressure curve of the pure solvent, at a
temperature lower than the freezing point of the pure solvent. (See Figure 14.8b) At this
point of intersection, the vapor pressure of the solution equals the vapor pressure of the
pure solvent.
39.
Water and ice are different phases of the same substance in equilibrium at the freezing
point of water, 0°C. The presence of the methanol lowers the vapor pressure and hence
the freezing point of water. If the ratio of alcohol to water is high, the freezing point can
be lowered as much as 10°C or more.
40.
Effectiveness in lowering the freezing point of 500. g water:
(a)
(b)
(c)
100. g (2.17 mol) of ethyl alcohol is more effective than 100. g (0.292 mol) of
sucrose.
20.0 g (0.435 mol) of ethyl alcohol is more effective than 100. g (0.292 mol) of
sucrose.
20.0 g (0.625 mol) of methyl alcohol is more effective than 20.0 g (0.435 mol) of
ethyl alcohol.
41.
Both molarity and molality describe the concentration of a solution. However, molarity is
the ratio of moles of solute per liter of solution, and molality is the ratio of moles of
solute per kilogram of solvent.
42.
5 molal NaCl = 5 mol NaCl>kg H 2O; 5 molar NaCl = 5 mol NaCl>L of solution. The
volume of the 5 molal solution will be larger than 1 liter (1 L H 2O + 5 mol NaCl). The
volume of the 5 molar solution is exactly 1 L (5 mol NaCl + sufficient H 2O to produce
1 L of solution). The molarity of a 5 molal solution is therefore, less than 5 molar.
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CHAPTER 14
SOLUTIONS TO EXERCISES
1.
Reasonably soluble: (a) KOH (b) NiCl 2
Insoluble: (c) ZnS
2.
Reasonably soluble: (c) CaCl 2 (d) Fe(NO3)3
Insoluble: (a) PbI 2 (b) MgCO3 (e) BaSO4
3.
Mass percent calculations
(a) 15.0 g KCl + 100.0 g H2O = 115.0 g solution
a
(b)
15.0 g
b11002 = 13.0% KCl
115.0 g
2.50 g Na3PO4 + 10.0 g H2O = 12.5 g solution
¢
(c)
(d) AgC2H 3O2
2.50 g
≤ 11002 = 20.0% Na3PO4
12.5 g
(0.20 mol NH4C2H3O2)a
77.09 g
b = 15 g NH4C2H3O2
1 mol
15 g NH4C2H3O2 + 125 g H2O = 140. g solution
¢
(d)
15 g
≤ 11002 = 11% NH4C2H3O2
140. g
(1.50 mol NaOH)a
40.00 g
b = 60.0 g NaOH
1 mol
18.02 g
b = 595 g H2O
1 mol
60.0 g NaOH + 595 g H2O = 655 g solution
(33.0 mol H2O)a
¢
4.
60.0 g
≤ 11002 = 9.16% NaOH
655 g
Mass percent calculations
(a) 25.0 g NaNO3 in 125.0 g H2O = 150.0
a
25.0 g
b11002 = 16.7% NaNO3
150.0 g
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(e) Na 2CrO4
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(b)
1.25 g CaCl2 in 35.0 g H2O = 36.3 g solution
¢
(c)
1.25 g
≤ 11002 = 3.44% CaCl2
36.3 g
(0.75 mol K2CrO4)a
194.2 g
b = 150 g K2CrO4
1 mol
150 g K2CrO4 + 225 g H2O = 380 g solution
a
(d)
150 g
b11002 = 39% K2CrO4
380 g
(1.20 mol H2SO4)a
(72.5 mol H2O)a
98.09 g
b = 118 g H2SO4
1 mol
18.02 g
b = 1.31 * 103 g H2O
1 mol
118 g H2SO4 + 1.31 * 103 g H2O = 1.43 * 103 g solution
a
5.
118 g
1.43 * 103 g
A 15.5% solution contains 15.5 g AgNO3 per 100. g solution
(25.2 g AgNO3)a
6.
100. g solution
b = 163 g solution
15.5 g AgNO3
A 10.0% NaCl solution contains 10.0 g NaCl per 100. g solution
(25.0 g NaCl) ¢
7.
b 11002 = 8.25% H2SO4
100. g solution
≤ = 250. g solution
10.0 g NaCl
(a)
(25 g solution) a
7.5 g CaSO4
b = 1.9 g CaSO4
100. g solution
(b)
(25 g solution)a
100. g solution - 7.5 g CaSO4
b = 23 g solvent
100. g solution
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8.
9.
(a)
(75 g solution) a
12.0 g BaCl2
b = 9.0 g BaCl2
100. g solution
(b)
(75 g solution) a
100. g solution - 12.0 g BaCl2
b = 66 g solvent
100. g solution
Mass/volume percent.
¢
10.
Mass/volume percent.
a
11.
10.0 mL CH 3OH
≤ 11002 = 25.0% CH 3OH
40.0 mL solution
Volume percent.
¢
13.
4.20 g NaCl
b11002 = 33.6% NaCl
12.5 mL solution
Volume percent.
¢
12.
22.0 g CH 3OH
≤ 11002 = 22.0% CH 3OH
100. mL solution
2.0 mL C6H 14
≤ 11002 = 22% C6H 14
9.0 mL solution
Molarity problems aM =
mol
b
L
(a)
a
0.25 mol 1000 mL
ba
b = 3.3 M
75.0 mL
1L
(b)
a
1.75 mol
b = 2.3 M KBr
0.75 L
(c)
¢
35.0 g
1 mol
b = 0.341 M NaC2H3O2
≤a
1.25 L 82.03 g
(d)
¢
75 g CuSO4• 5 H2O
1 mol
b = 0.30 M CuSO4
≤a
1.0 L
249.7 g
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14.
15.
Molarity problems aM =
(a)
a
0.50 mol 1000 mL
ba
b = 4.0 M
125 mL
1L
(b)
a
2.25 mol
b = 1.50 M CaCl2
1.50 L
(c)
¢
275 g
1 mol
1000 mL
ba
b = 1.97 M C6H12O6
≤a
775 mL 180.2 g
1L
(d)
¢
125 g MgSO4• 7 H2O
1 mol
b = 0.203 M MgSO4
≤a
2.50 L
246.5 g
(b)
(c)
mol solute
or mol solute = 1L solution21Molarity2
L solution
1.50 mol HNO3
10.75 L2a
b = 1.1 mol HNO3
L
0.75 mol NaClO3
1L
110.0 mL2a
ba
b = 7.5 * 10 - 3 mol NaClO3
1000 mL
L
1L
0.50 mol LiBr
1175 mL2a
ba
b = 0.088 mol LiBr
1000 mL
L
Molarity =
(a)
(b)
(c)
17.
mol solute
or mol solute = 1L solution21Molarity2
L solution
1.20 mol H2SO4
11.5 L2a
b = 1.8 mol H2SO4
L
0.0015 mol BaCl2
1L
125.0 mL2a
ba
b = 3.8 * 10 - 5 mol BaCl2
1000 mL
L
0.35 mol K3PO4
1L
1125 mL2a
ba
b = 0.044 mol K3PO4
1000 mL
L
Molarity =
(a)
16.
mol
b
L
(a)
12.5 L2a
0.75 mol K2CrO4 194.2 g
ba
b = 360 g K2CrO4
L
1 mol
(b)
175.2 mL2a
(c)
1250 mL2a
0.050 mol HC2H3O2 60.05 g
1L
ba
ba
b = 0.226 g HC2H3O2
1000 mL
L
1 mol
16 mol HNO3 63.02 g
1L
ba
ba
b = 250 g HNO3
1000 mL
L
1 mol
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18.
19.
20.
21.
(a)
11.20 L2a
(b)
127.5 mL2 ¢
(c)
1120 mL2 ¢
(a)
10.15 mol H3PO42a
(b)
135.5 g H3PO42a
(a)
10.85 mol NH4Cl2a
(b)
125.2 g) NH4Cla
18 mol H2SO4 98.09 g
ba
b = 2.1 * 103 g H2SO4
L
1 mol
1.5 mol KMnO4 158.0 g
1L
ba
b = 6.52 g KMnO4
≤a
1000 mL
L
1 mol
0.025 mol Fe2(SO4)3 399.9 g
1L
ba
b = 1.2 g Fe2(SO4)3
≤a
1000 mL
L
1 mol
1000 mL
1L
ba
b = 2.0 * 102 mL
0.750 mol H3PO4
1L
1 mol
1L
1000 mL
ba
ba
b = 483 mL
97.99 g
0.750 mol H3PO4
1L
1000 mL
1L
ba
b = 3.4 * 103 mL
0.250 mol NH4Cl
1L
1 mol
1L
1000 mL
ba
ba
b = 1.88 * 103 mL
53.49 g 0.250 mol NH4Cl
L
Dilution problem V1M 1 = V2M 2
(a)
V1 = 125 mL
V2 = (125 mL + 775 mL) = 900. mL
M1 = 5.0 M
M2 = M2
1125 mL215.0 M2 = 1900. mL21M22
M2 =
(b)
1125 mL215.0 M2
= 0.694 M
900. mL
V1 = 250 mL
V2 = (250 mL + 750 mL) = 1.00 * 103 mL
M1 = 0.25 M
M2 = M2
1250 mL210.25 M2 = 11.00 * 103 mL21M22
1250 mL210.25 M2
= 0.063 M
M2 =
1.00 * 103 mL
(c)
First calculate the moles of HNO3 in each solution. Then calculate the molarity.
1L
0.50 mol
(75 mL)a
ba
b = 0.038 mol HNO3
1000 mL
1L
1L
1.5 mol
(75 mL)a
ba
b = 0.11 mol HNO3
1000 mL
1L
Total mol = 0.15 mol
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Total volume = 75 mL + 75 mL = 150. mL = 0.150 L
0.150 mol
= 1.00 M
0.150 L
22.
Dilution problem V1M 1 = V2M 2
(a)
V1 = 175 mL
V2 = (175 mL + 275 mL) = 450. mL
M1 = 3.0 M
M2 = M2
1175 mL213.0 M2 = 1450. mL21M22
M2 =
(b)
1175 mL213.0 M2
= 1.2 M
450. mL
V1 = 350 mL
M1 = 0.10 M
V2 = (350 mL + 150 mL) = 5.0 * 102 mL
M2 = M2
1350 mL210.10 M2 = 15.0 * 102 mL21M22
M2 =
(c)
1350 mL210.10 M2
5.0 * 102 mL
= 0.070 M
First calculate the moles of HCl in each solution. Then calculate the molarity.
1L
0.250 mol
ba
b = 0.0125 mol HCl
1000 mL
1L
1L
0.500 mol
(25.0 mL)a
ba
b = 0.0125 mol HCl
1000 mL
1L
Total mol = 0.0250 mol
Total volume = 50.0 mL + 25.0 mL = 75.0 mL = 0.0750 L
0.250 mol
= 0.333 M
0.0750 L
(50.0 mL)a
23.
V1M 1 = V2M 2
(a)
(b)
1V12115 M2 = 1750 mL213.0 M2
1750 mL213.0 M2
V1 =
= 150 mL 15 M H3PO4
15 M
1V12116 M2 = 1250 mL210.50 M2
V1 =
1250 mL210.50 M2
= 7.8 mL 16 M HNO3
16 M
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24.
V1M 1 = V2M 2
(a)
(b)
25.
1V12118 M2 = 1225 mL212.0 M2
1225 mL212.0 M2
V1 =
= 25 mL 18 M H2SO4
18 M
1V12115 M2 = 175 mL211.0 M2
175 mL211.0 M2
V1 =
= 5.0 mL 15 M NH3
15 M
10.125 L2a
(a)
6.0 mol HC2H3O2
b = 0.75 mol HC2H3O2
L
Final volume after mixing
125 mL + 525 mL = 650. mL = 0.650 L
0.75 mol HC2H3O2
= 1.2 M HC2H3O2
0.650 L
(b)
1175 mL2a
1.5 mol HC2H3O2
1L
b¢
≤ = 0.26 mol HC2H3O2
1000 mL
L
Total moles = 0.75 mol + 0.26 mol = 1.01 mol HC2H3O2
Final volume = 125 mL + 175 mL = 300. mL = 0.300 L
1.01 mol HC2H3O2
= 3.37 M HC2H3O2
0.300 L
26.
10.175 L2a
(a)
(b)
3.0 mol HCl
b = 0.53 mol HCl
L
Final volume after mixing
175 mL + 250 mL = 425 mL = 0.425 L
0.53 mol HCl
= 1.2 M HCl
0.425 L
1115 mL2a
6.0 mol HCl
1L
b¢
≤ = 0.69 mol HCl
1000 mL
L
Total moles = 0.53 mol + 0.69 mol = 1.22 mol HCll
Final volume = 175 mL + 115 mL = 290. mL = 0.290 L
1.22 mol HCl
= 4.21 M HCl
0.290 L
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27.
3 Ca(NO3)2(aq) + 2 Na3PO4(aq) ¡ Ca3(PO4)2(s) + 6 NaNO3(aq),
(a)
12.7 mol Na3PO42 ¢
(b)
10.75 mol Ca(NO3)22 ¢
(c)
L Ca(NO3)2 ¡ mol Ca(NO3)2 ¡ mol Na3PO4
1 mol Ca3 (PO4)2
≤ = 1.4 mol Ca3(PO4)2
2 mol Na3PO4
11.45 L Ca(NO3)22 ¢
(d)
2 mol Na3PO4
0.225 mol
b = 0.218 mol Na3PO4
≤a
L
3 mol Ca(NO3)2
mL Ca(NO3)2 ¡ mol Ca(NO3)2 ¡ mol Ca3(PO4)2 ¡ g Ca3(PO4)2
1125 mL Ca(NO3)22 ¢
(e)
0.500 mol 1 mol Ca3(PO4)2 310.18 g
ba
b = 6.46 g Ca3(PO4)2
≤a
1000 mL
3 mol Ca(NO3)2
mol
mL Ca(NO3)2 ¡ mol Ca(NO3)2 ¡ mol Na3PO4 ¡ mL Na3PO4
115.0 mL Ca(NO3)22 ¢
(f)
2 mol Na3PO4
0.50 mol
1000 mL
ba
b = 20. mL Na3PO4
≤a
1000 mL 3 mol Ca(NO3)2
0.25 mol
Find mol Ca(NO3)2 mL Na3PO4 ¡ mol Na3PO4 ¡ mol Ca(NO3)2
150.0 mL Na3PO4) ¢
M =
28.
6 mol NaNO3
≤ = 1.5 mol NaNO3
3 mol Ca(NO3)2
mol
L
3 mol Ca(NO3)2
2.0 mol
b = 0.15 mol Ca(NO3)2
≤a
1000 mL
2 mol Na3PO4
M = a
0.15 mol Ca(NO3)2
b = 3.0 M Ca(NO3)2
0.0500 L
2 NaOH(aq) + H2SO4(aq) ¡ Na2SO4(aq) + 2 H2O(l),
(a)
13.6 mol H2SO42 ¢
(b)
10.025 mol NaOH2 ¢
(c)
L H2SO4 ¡ mol H2SO4 ¡ mol NaOH
12.50 L H2SO42 ¢
1 mol Na2SO4
≤ = 3.6 mol Na2SO4
1 mol H2SO4
2 mol H2O
≤ = 0.025 mol H2O
2 mol NaOH
0.125 mol
2 mol NaOH
b = 0.625 mol NaOH
≤a
1L
1 mol H2SO4
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(d)
mL NaOH ¡ mol NaOH ¡ mol Na2SO4 ¡ g Na2SO4
125 mL NaOH2 ¢
(e)
0.050 mol 1 mol Na2SO4 142.05 g
ba
b = 0.089 g Na2SO4
≤a
1000 mL
2 mol NaOH
mol
mL NaOH ¡ mol NaOH ¡ mol H2SO4 ¡ mL H2SO4
125.5 mL NaOH2 ¢
(f)
Find mol NaOH
0.750 mol 1 mol H2SO4
1000 mL
ba
b = 38.25 mL H2SO4
≤a
1000 mL
2 mol NaOH
0.250 mol
mL H2SO4 ¡ mol H2SO4 ¡ mol NaOH
135.72 mL H2SO4) ¢
M =
29.
mol
L
0.125 mol
2 mol NaOH
b = 8.93 * 10 - 3 mol NaOH
≤a
1000 mL
1 mol H2SO4
M = a
8.93 * 10 - 3 mol NaOH
b = 0.185 M NaOH
0.04820 L
2 KMnO4(aq) + 16 HCl(aq) ¡ 2 MnCl2(aq) + 5 Cl2(g) + 8 H2O(l) + 2 KCl(aq)
(a)
115.0 mL HCl2 ¢
(b)
11.85 mol MnCl22a
(c)
1125 mL KCl2 ¢
(d)
(15.60 mL KMnO4)a
M = a
(e)
2 mol KMnO4
1L
ba
b = 12.3 L KMnO4
2 mol MnCl2
0.150 mol KMnO4
0.525 mol 16 mol HCl 1000 mL
ba
b = 210. mL HCl
≤a
1000 mL
2 mol KCl
2.50 mol
0.250 mol
16 mol HCl
b = 0.0312 mol HCl
ba
1000 mL
2 mol KMnO4
0.0312 mol HCl
b = 1.41 M HCl
0.02220 L
mL HCl ¡ mol HCl ¡ mol Cl2 ¡ L Cl2 (gas at STP)
1125 mL HCl2 ¢
(f)
8 mol H2O
0.250 mol
b = 1.88 * 10 - 3 mol H2O
≤a
1000 mL
16 mol HCl
5 mol Cl2
2.5 mol
22.4 L
ba
b = 2.2 L Cl2
≤a
1000 mL 16 mol HCl
mol
Limiting reactant problem. Convert volume of both reactants to liters of Cl2 gas.
115.0 mL HCl) a
5 mol Cl2
22.4 L
0.750 mol
ba
ba
b = 0.0788 L Cl2
1000 mL
16 mol HCl
mol
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112.0 mL KMnO4)a
5 mol Cl2
0.550 mol
22.4 L
ba
ba
b = 0.373 L Cl2
1000 mL
2 mol KMnO4
mol
HCl is the limiting reactant. 0.0788 L of Cl2 are produced.
30.
K2CO3(aq) + 2 HC2H3O2(aq) ¡ 2 KC2H3O2(aq) + H2O(l) + CO2(g)
(a)
125.0 mL HC2H3O22 ¢
1 mol H2O
0.150 mol
b = 1.88 * 10 - 3 mol H2O
≤a
1000 mL
2 mol HC2H3O2
(b)
117.5 mol KC2H3O22a
(c)
175.2 mL K2CO32 ¢
1 mol K2CO3
1L
ba
b = 41.7 L K2CO3
2 mol KC2H3O2 0.210 mol K2CO3
0.750 mol 2 mol HC2H3O2
1000 mL
ba
b
≤a
1000 mL
1 mol K2CO3
1.25 mol HC2H3O2
= 90.2 mL HC2H3O2
(d)
(18.50 mL K2CO3)a
M = a
(e)
0.250 mol 2 mol HC2H3O2
b = 9.25 * 10 - 3 mol HC2H3O2
ba
1000 mL
1 mol K2CO3
9.25 * 10 - 3 mol HC2H3O2
b = 0.911 M HC2H3O2
0.01015 L
mL HCl ¡ mol HCl ¡ mol Cl2 ¡ L Cl2 (gas at STP)
1105 mL of HC2H3O22 ¢
(f)
1 mol CO2
1.5 mol
22.4 L
ba
b = 1.8 L CO2
≤a
1000 mL 2 mol HC2H3O2
mol
Limiting reactant problem. Convert volume of both reactant to liters of CO2 gas.
1 mol CO2
0.350 mol
22.4 L
125.0 mL K2CO3)a
ba
ba
b = 0.196 L CO2
1000 mL
1 mol K2CO3
mol
125.0 mL HC2H3O2)a
1 mol CO2
0.250 mol
22.4 L
ba
ba
b = 0.0700 L CO2
1000 mL
2 mol HC2H3O2
mol
HC2H3O2 is the limiting reactant. 0.0700 L of CO2 are produced.
31.
Molality = m =
mol solute
kg solvent
(a)
¢
14.0 g CH 3OH 1000 g
4.37 mol CH 3OH
1 mol
ba
b = ¢
≤a
≤ = 4.37 m CH 3OH
100. g H 2O
kg
32.04 g
kg H 2O
(b)
¢
2.50 mol C6H 6 1000 g
10. mol C6H 6
b = ¢
≤a
≤ = 10. m C6H 6
250 g C6H 14
kg
kg C6H 14
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32.
33.
Molality = m =
mol solute
kg solvent
(a)
¢
1.0 g C6H 12O6 1000 g
5.5 mol C6H 12O6
1 mol
ba
b = ¢
≤a
≤ = 5.5 m C6H 12O6
1.0 g H 2O
kg
180.2 g
kg H 2O
(b)
¢
0.250 g I2
1 mol I2
b = 9.9 * 10 - 4 m I2
≤a
1.0 kg H2O 253.8 g I2
(a)
¢
2.68 g C10H 8
1000 g C6H 6
1 mol
≤¢
≤¢
≤ = 0.544 m
38.4 g C6H 6
128.2 g C10H 8
kg
(b)
K f (for benzene) =
5.1°C
m
¢t f = 10.544 m2a
Freezing point of benzene = 5.5°C
5.1°C
b = 2.8°C
m
Freezing point of solution = 5.5°C - 2.8°C = 2.7°C
(c)
K b (for benzene) =
¢t b = (0.544 m)a
2.53°C
m
Boiling point of benzene = 80.1°C
2.53°C
b = 1.38°C
m
Boiling point of solution = 80.1°C + 1.38°C = 81.5°C
34.
100.0 g C2H 6O2
1000 g
1 mol
ba
b = 10.74 m
≤a
150.0 g H 2O
62.07 g
kg
(a)
¢
(b)
¢t b = mK b = (10.74 m)a
0.512°C
b = 5.50°C (Increase in boiling point)
m
Boiling point = 100.00°C + 5.50°C = 105.50°C
(c)
¢t f = mK f = 110.74 m2a
1.86°C
b = 20.0°C (Decrease in freezing point)
m
Freezing point = 0.00°C - 20.0°C = -20.0°C
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35.
Freezing point of acetic acid is 16.6°C K f acetic acid =
3.90°C
m
¢t f = 16.6°C - 13.2°C = 3.4°C
¢t f = mK f
m =
3.4°C
= 0.87 m
3.90°C>m
Convert 8.00 g unknown> 60.0 g HC2H 3O2 to g> mol (molar mass)
Conversion:
¢
36.
g unknown
g
g unknown
¡
¡
g HC2H 3O2
kg HC2H 3O2
mol
8.00 g unknown
1 kg HC2H 3O2
1000 g
b¢
≤a
≤ = 153 g>mol
60.0 g HC2H 3O2
kg
0.87 mol unknown
¢t f = 2.50°C
K f (for H 2O) =
1.86°C
m
¢t f = mK f
2.50°C
m =
= 1.34 m
1.86°C>m
Convert 4.80 g unknown> 22.0 g H 2O to g> mol (molar mass)
¢
4.80 g unknown 1000 g
1 kg H 2O
b¢
≤a
≤ = 163 g>mol
22.0 g H 2O
kg
1.34 mol unknown
37.
Salt (NaCl) is an ionic compound. When it is dissolved in water the sodium and
chloride ions separate or dissociate. The polar water molecules are attracted to the
polar sodium and chloride ions and are hydrated (surrounded by water molecules).
The sodium and chloride ions are separated from one another and distributed
throughout the water in this way. Na+ and Cl- are hydrated in aqueous solution: See
Question 1.
38.
Sugar molecules are not ionic; therefore they do not dissociate when dissolved in water.
However, sugar molecules are polar, so water molecules are attracted to them and the
sugar becomes hydrated. The water molecules help separate the sugar molecules from
each other and distribute them throughout the water.
39.
Sugar and salt behave differently when dissolved in water because salt is an ionic
compound and sugar is a molecular compound.
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40.
An isotonic sodium chloride solution has the same osmotic pressure as human blood
plasma. When blood cells are placed in an isotonic solution the osmotic pressure inside
the cells is equal to the osmotic pressure outside the cells so there is no change in the
appearance of the blood cells.
41.
The KMnO4 crystals give the solution its purple color. The purple streaks are formed
because the solute has not been evenly distributed throughout that solvent yet. The
MnO4 - has a purple color in solution.
42.
The line for KNO3 slopes upward, because the solubility increases as the temperature
increases. KNO3 has the steepest slope of all the compounds given in the diagram. It
exhibits the greatest increase in the number of grams of solute that is able to dissolve
in 100 g of water than any other compound in the diagram as the temperature increases.
43.
First calculate the g NaOH to neutralize the HCl.
NaOH + HCl ¡ NaCl + H 2O
1.0 mol 1 mol NaOH 40.00 g
ba
ba
b = 6.0 g NaOH required to neutralize
10.15 L HCl2a
L
1 mol HCl
mol
the acid
Now calculate the grams of 10% NaOH solution that contains 6.0 g NaOH
6.0 g NaOH
10.0 g NaOH
=
x
100.0 g 10.0% NaOH solution
x = 60. g 10% NaOH solution
44.
1.0 m HCl =
36.46 g HCl
1 mol HCl
=
1 kg H 2O
1000 g H 2O
Total mass of solution = 1000 g + 36.46 g = 1036.46 g
Therefore, 1.0 m HCl =
1 mol HCl
1036.46 g HCl solution
NaOH + HCl ¡ NaCl + H 2O
Calculate the grams NaOH to neutralize HCl
1250.0 g solution2a
1 mol NaOH 40.00 g
1 mol HCl
ba
ba
b = 9.648 g NaOH
1036.46 solution
1 mol HCl
mol
Calculate the grams of 10.0% NaOH solution that contains 9.648 g NaOH.
10.0 g NaOH
9.648 g NaOH
=
x
100.0 g 10.0% NaOH solution
x = 96.5 g 10% NaOH solution
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45.
(a)
11.0 L syrup2a
(b)
¢
(c)
m =
¢
46.
Kf =
¢
15.0 g sugar
1000 mL 1.06 g
ba
ba
b = 1.6 * 102 g sugar
L
mL
100. g syrup
1.6 * 102 g C12H 22O11
1 mol
b = 0.47 M
≤a
L
342.3 g
mol sugar
kg H 2O
15% sugar by mass = 15.0 g C12H 22O11 + 85.0 g H 2O
15.0 g C12H 22O11 1000 g H 2O
1 mol
≤¢
≤¢
≤ = 0.516 m
85.0 g H 2O
1 kg H 2O
342.3 g C12H 22O11
5.1°C
m
¢t f = 0.614°C
3.84 g C4H 2N 1000 g
15.4 g C4H 2N
b =
≤a
250.0 g C6H 6
kg
kg C6H 6
¢t f = mK f
m =
¢
0.12 mol C4H 2N
0.614°C
= 0.12 m =
5.1°C>m
kg C6H 6
15.4 g C4H 2N
1 kg C6H 6
≤¢
≤ = 128 g>mol = 1.3 * 102 g>mol
kg C6H 6
0.12 mol C4H 2N
Empirical mass (C4H 2N) = 64.07 g
130 g
= 2.0 (number of empirical formulas per molecular formula)
64.07 g
Therefore, the molecular formula is twice the empirical formula, or C8H 4N2 .
47.
112.0 mol HCl2a
11.00 L2a
36.46 g
b = 438 g HCl in 1.00 L solution
mol
1.18 g solution 1000 mL
ba
b = 1180 g solution
mL
L
1180 g solution - 438 g HCl = 742 g H 2O (0.742 kg H 2O)
Since molality =
48.
mol HCl
12.0 mol HCl
=
= 16.2 m HCl
kg H 2O
0.742 kg H 2O
First calculate the g KNO3 in the solution.
The conversion is:
mg K +
g K+
g KNO3
¡
¡
¡ g KNO3
mL
mL
mL
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¢
5.5 mg K +
101.1 g KNO3
1g
b¢
≤a
≤ 1450 mL2 = 6.4 g KNO3
mL
1000 mg
39.10 g K +
Now calculate the mol KNO3 and the molarity.
16.4 g KNO32a
1 mol
b = 0.063 mol KNO3
101.1 g
0.063 mol KNO3
= 0.14 M
0.450 L
49.
125.0 g KCl2a
100. g solution
b = 455 g solution
5.50 g KCl
Alternate solution:
a
25.0 g KCl
5.50 g KCl
b
b = a
x
100. g solution
x = 455 g solution
50.
Verification of K b for water
¢t b = mK b
¢t b = 101.62°C - 100°C = 1.62°C
Kb =
¢t b
m
First calculate the molality of the solution.
16.10 g C2H 6O2
3.16 mol C2H 6O2
=
162.07 g>mol210.0820 kg H 2O2
kg H 2O
m =
0.513°C kg H 2O
¢t b
1.62°C
=
=
m
3.16 mol>kg H 2O
mol
Kb =
51.
(a)
1500.0 mL solution2a
(b)
a
0.90 g NaCl
b = 4.5 g NaCl
100. mL solution
4.5 g NaCl
b11002 = 9.0%
x mL
x =
x = volume of 9.0% solution
4.5 g NaCl
= 50. mL (4.5 g NaCl in solution)
9.0%
500. mL - 50. mL = 450. mL H 2O must evaporate
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52.
From Figure 14.4, the solubility of KNO3 in H 2O at 20°C is 32 g per 100. g H 2O.
150.0 g KNO32a
100. g H 2O
b = 156 g H 2O to produce a saturated solution.
32.0 g KNO3
175 g H 2O - 156 g H 2O = 19 g H 2O must be evaporated.
53.
1150 mL alcohol2a
54.
(a)
11.00 L solution2a
(b)
1500. g HNO32 ¢
55.
35.0 g HNO3
1000 mL solution 1.21 g
ba
b¢
≤ = 424 g HNO3
L solution
mL
100. g solution
1000 mL solution
1.00 L
b = 1.18 L solution
≤a
424 g HNO3
1000 mL
Assume 1.000 L (1000. mL) of solution
a
56.
100. mL solution
b = 210 mL solution
70.0 mL alcohol
35.0 g HNO3
1000. mL 1.21 g solution
1 mol
ba
b¢
b = 6.72 M HNO3
≤a
L
mL
100. g solution 63.02 g
First calculate the molarity of the solution
¢
80.0 g H 2SO4 1000 mL
1 mol
ba
b = 1.63 M H 2SO4
≤a
500. mL
L
98.09 g
M 1V1 = M 2V2
11.63 M21500. mL2 = 10.10 M21V22
V2 =
57.
M 1V1 = M 2V2
116 M2110.0 mL2 = 1M 221500. mL2
M2 =
58.
11.63 M21500. mL2
= 8.2 * 103 mL = 8.2 L
0.10 M
116 M2110.0 mL2
= 0.32 M HNO3
500.0 M
Mg + 2 HCl ¡ MgCl 2 + H 2(g)
(a)
mL HCl ¡ mol HCl ¡ mol H 2
1200.0 mL HCl2a
1 mol H 2
3.00 mol
b¢
≤ = 0.300 mol H 2
1000 mL 2 mol HCI
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(b)
PV = nRT
P = 1720 torr2a
1 atm
b = 0.95 atm
760 torr
T = 27°C = 300. K
n = 0.300 mol
V =
59.
10.300 mol210.0821 L atm>mol K21300. K2
nRT
=
= 7.8 L H 2
P
0.95 atm
Mg(OH)2 + 2 HCl ¡ MgCl 2 + 2 H 2O
Al(OH)3 + 3 HCl ¡ AlCl 3 + 3 H 2O
Calculate the moles of HCl neutralized by each base.
11.20 g Mg(OH)22a
11.00 g Al(OH)32a
1 mol
2 mol HCl
b¢
≤ = 0.0411 mol HCl
58.33 g 1 mol Mg(OH)2
3 mol HCl
1 mol
b¢
≤ = 0.0385 mol HCl
78.00 g 1 mol Al(OH)3
1.20 g Mg(OH)2 reacts with more HCl than 1.00 g Al(OH)3 . Therefore, Mg(OH)2 is
more effective in neutralizing stomach acid.
60.
61.
(a)
With equal masses of CH 3OH and C2H 5OH, the substance with the lower molar
mass will represent more moles of solute in solution. Therefore, the CH 3OH will be
more effective than C2H 5OH as an antifreeze.
(b)
Equal molal solutions will lower the freezing point of the solution by the same
amount.
Calculate molarity and molality. Assume 1000 mL of solution to calculate the amounts of
H 2SO4 and H 2O in the solution.
11000 mL solution2a
1.29 g
b = 1.29 * 103 g solution
mL
11.29 * 103 g solution2 ¢
38 g H 2SO4
≤ = 4.9 * 102 g H 2SO4
100 g solution
1.29 * 103 g solution - 4.9 * 102 g H 2SO4 = 8.0 * 102 g H 2O in the solution
m = ¢
490 g H 2SO4
2
8.0 * 10 g H 2O
≤a
1000 g
1 mol
ba
b = 6.2 m H 2SO4
kg
98.09 g
4.9 * 102 g H 2SO4
1 mol
b = 5.0 M H 2SO4
M = ¢
≤a
L
98.09 g
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62.
Freezing point depression is 5.4°C
(a)
¢t f = mK f
¢t f
5.4°C
=
= 2.9 m
Kf
1.86°C kg solvent>mol solute
m =
(b)
K b (for H 2O) =
0.512°C kg solvent
0.512°C
=
m
mol solute
¢t b = mK b = 12.9 m2a
0.512°C
b = 1.5°C
m
Boiling point = 100°C + 1.5°C = 101.5°C
63.
Freezing point depression = 0.372°C
Kf =
1.86°C
m
¢t f = mK f
m =
0.372°C
= 0.200 m
1.86°C>m
16.20 g C2H 6O22a
1 mol
b = 0.100 mol C2H 6O2
62.07 g
10.100 mol C2H 6O22 ¢
64.
(a)
1 kg H 2O
1000 g H 2O
≤¢
≤ = 500. g H 2O
0.200 mol C2H 6O2
kg H 2O
Freezing point depression = 20.0°C
12.0 L H 2Oa
1000 mL 1.00 g
ba
b = 1.20 * 104 g H 2O
L
mL
¢t f = mK f
m =
20.0°C
= 10.8 m
1.86°C>m
11.20 * 104 g H 2O2 ¢
10.8 mol C2H 6O2 62.07 g
b = 8.04 * 103 g C2H 6O2
≤a
1000 g H 2O
mol
(b)
18.04 * 103 g C2H 6O22a
1.00 mL
b = 7.24 * 103 mL C2H 6O2
1.11 g
(c)
°F = 1.8 (°C) + 32 = 1.8 (-20.0) + 32 = -4.0°F
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65.
HNO3 + NaHCO3 ¡ NaNO3 + H 2O + CO2
First calculate the grams of NaHCO3 in the sample.
mL HNO3 ¡ L HNO3 ¡ mol HNO3 ¡ mol NaHCO3 ¡ g NaHCO3
1150 mL HNO32a
1L
0.055 mol 1 mol NaHCO3 84.01 g
ba
b¢
b
≤a
1000 mL
L
1 mol HNO3
mol
= 0.69 g NaHCO3 in the sample
¢
66.
0.69 g NaHCO3
≤ 11002 = 47% NaHCO3
1.48 g sample
(a)
Dilution problem: M 1V1 = M 2V2
11.5 M218.4 L2 = 117.8 M21V22
V2 =
11.5 M218.4 L2
= 0.71 L
17.8 M
0.71 L of 17.8 M H 2SO4 is to be diluted to 8.4 L.
8.4 L - 0.71 L = 7.7 L H 2O must be added (assume volumes are additive)
67.
(b)
a
17.8 mol
b11.00 mL2 = 0.0178 mol
1000. mL
(c)
a
1.5 mol
b11.00 mL2 = 0.0015 mol H 2SO4 in each mL
1000. mL
moles HNO3 total = moles HNO3 from 3.00 M + moles HNO3 from 12.0 M
M TVT = M 3.00 MV3.00 M + M 12.0 MV12.0 M
Assume preparation of 1000. mL of 6 M solution
Let y = volume of 3.00 M solution; volume of 12.0 M = 1000. mL - y
16.00 M211000. mL2 = 13.00 M21y2 + 112.0 M211000. mL - y2
6000. mL = 3.00 y mL + 12,000 mL - 12.0 y
6000. mL = 9.00 y
y =
6000. mL
= 667 mL 3 M
9.00
1000. mL - 667 mL = 333 mL 12 M
Mix together 667 mL 3.00 M HNO3 and 333 mL of 12.0 M HNO3 to get 1000. mL
of 6.00 M HNO3 .
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68.
HBr + NaOH ¡ NaBr + H 2O
First calculate the molarity of the diluted HBr solution.
The reaction is 1 mol HBr to 1 mol NaOH, so
M AVA = M BVB
1M A21100.0 mL2 = 10.37 M2188.4 mL2
10.37 M2188.4 mL2
= 0.33 M HBr (diluted solution)
100.00 mL
MA =
Now calculate the molarity of the HBr before dilution.
M 1V1 = M 2V2
1M 12120.0 mL2 = 10.33 M21240. mL2
M1 =
69.
10.33 M21240. mL2
= 4.0 M HBr (original solution)
20.0 mL
Ba(NO3)2 + 2 KOH ¡ Ba(OH)2 + 2 KNO3
This is a limiting reactant problem. First calculate the moles of each reactant and
determine the limiting reactant.
M * L = a
moles
b1L2 = moles
L
a
0.642 mol
b10.0805 L2 = 0.0517 mol Ba(NO3)2
L
a
0.743 mol
b10.0445 L2 = 0.0331 mol KOH
L
According to the equation, twice as many moles of KOH as Ba(NO3)2 are needed, so
KOH is the limiting reactant.
10.0331 mol KOH2 ¢
70.
1 mol Ba(OH)2 171.3 g
b = 2.84 g Ba(OH)2 is formed
≤a
2 mol KOH
mol
(a)
a
0.25 mol
b10.0458 L2 = 0.011 mol Li 2CO3
L
(b)
a
73.89 g
0.25 mol
b10.75 L2a
b = 14 g Li 2CO3
L
mol
(c)
16.0 g Li 2CO32a
1 mol
1000. mL
ba
b = 3.2 * 102 mL solution
73.89 g
0.25 mol
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HEINS14-180-206v3.qxd
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Page 206
- Chapter 14 -
(d)
Assume 1000. mL solution
a
1.22 g
b11000. mL2 = 1220 g solution
mL
a
0.25 mol 73.89 g Li 2CO3
b¢
≤ = 18 g Li 2CO3 per L solution
L
mol
% = a
71.
18 g
g solute
b11002 = a
b11002 = 1.5%
g solution
1220 g
The balanced equation is
2 HCl(aq) + Na2SO3(aq) ¡ 2 NaCL(aq) + H2O(l) + SO2(g)
(125 mL HCl)a
1 mol SO2
2.50 mol
ba
b = 0.156 mol SO2
1000 mL 2 mol HCl
1 mol SO2
1.75 mol
ba
b = 0.131 mol SO2
1000 mL 1 mol Na2SO3
Na2SO3 is the limiting reactant. 0.131 mol of SO2 gas will be produced
(75.0 mL Na2SO3)a
The gas is at non-standard conditions, so use PV = nRT to find the liters of SO2.
V =
72.
nRT
P
V =
(0.131 mol)(0.0821 L atm/mol K)(295 K)
= 3.11 L SO2
(775 torr)(1 atm/760 torr)
mass of solute = mass of container & solute - mass of container - mass of water
mass of water = 15.549 moles2118.02 g>mol2 = 100.0 g
mass of solute = 563 g - 375 g - 100.0 g = 88 g
solubility in water = g solute>100 g H 2O
Using this data, solubility is 88 g solute>100.0 g water = NaNO3 (see Table 14.3).
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