Chemistry 30AP
Chapter 13 (mostly) - Properties of Solutions
1. Create a vocabulary list from the terms in bold on pages 563 and 564. Include the terms solute, solvent,
solubility, dissolution, precipitate, electrolyte, nonelectrolyte, spectator ions, polar, nonpolar.
2. Write dissociation equations for the substances in question 4.16. Determine the concentration of each ion in
solution if each salt has molarity of 0.20 mol/L.
a) MgI2 (aq)  Mg2+(aq) + 2 I1-(aq)
0.20 M
0.40 M
b) Al(NO3)3 (aq)  Al3+(aq) + 3 NO31-(aq)
0.20 M
0.60 M
c) HClO3 (aq)  H1+(aq) + ClO41-(aq)
0.20 M
0.20 M
d) KC2H3O2 (aq)  K1+(aq) + C2H3O21-(aq)
0.20 M
0.20 M
3. Predict the solubility of the substances found in questions 4.19 and 4.20.
4. Write a molecular equation, complete ionic equation and net ionic equation for 4.24.
a) Cr2(SO4)3 (aq)
+
3 (NH4)2CO3 (aq)  Cr2(CO3)3 (s) + 3 (NH4)2SO4 (aq)
2 Cr3+(aq) + 3 SO42- (aq) + 6 NH41+(aq) + 3 CO32- (aq)  Cr2(CO3)3 (s) + 6 NH41+(aq) + 3 SO42- (aq)
2 Cr3+(aq) + 3 CO32- (aq)  Cr2(CO3)3 (s)
b) Ba(NO3)2 (aq)
+
K2SO4 (aq)  BaSO4 (s) + 2 KNO3 (aq)
Ba2+(aq) + 2 NO31- (aq) + 2 K1+(aq) + SO42- (aq)  BaSO4 (s) + 2 NO31- (aq) + 2 K1+(aq)
Ba2+(aq) + SO42- (aq)  BaSO4 (s)
c) Fe(NO3)3 (aq)
+
3 KOH (aq)  Fe(OH)3 (s) + 3 KNO3 (aq)
Fe3+(aq) + 3 NO31- (aq) + 3 K1+(aq) + 3 OH1- (aq)  Fe(OH)3 (s) + 3 NO31- (aq) + 3 K1+(aq)
Fe3+(aq) + 3 OH1- (aq)  Fe(OH)3 (s)
5. Draw Lewis structures for the substances in question 8.46.
6. Draw Lewis structures, give the electron-domain and molecular geometries, and determine the polarity of the
substances in question 9.36.
a) IF
-
b) CS2
- 16 valence electrons; C does not have octet after each S gets 8 electrons,
so double bonds are made.
- C has linear electron domain
- molecular geometry is linear
- bond is polar covalent, but symmetry causes bond dipoles to cancel; the
molecule is non-polar
c) SO3
-
24 valence electrons; 1 double bond is made to give S a full octet.
S has trigonal planar electron domain
molecular geometry is planar trigonal
bond is slightly polar covalent, but symmetry causes bond dipoles to cancel;
the molecule is non-polar
d) PCl3
-
26 valence electrons.
P has tetrahedral electron domain
molecular geometry is trigonal pyramidal
P-Cl bond is polar and the shape means the dipoles do not cancel; this is a
polar molecule
e) SF6
- 48 valence electrons
- electron domain and molecular geometry are octahedral
- S-F bond is polar, but the symmetry means that the bond dipoles cancel. This
is a non-polar molecule.
f)
-
IF5
14 valence electrons
both have a tetrahedral electron domain
since each makes 1 bond, the molecular geometry is “linear”
the bond is polar covalent, since the atoms are well separated on the
periodic table (large electronegativity difference). There is no symmetry to
cancel the bond dipole, so this molecule is polar.
42 valence electrons
electron domain is octahedral
molecular geometry is square pyramidal
I-F bond is polar, but the symmetry means that the bond dipoles do not
cancel. This is a polar molecule.
7. Explain briefly how the solution process works, including enthalpy and entropy considerations (13.1)
8. Use the principle of “like dissolves like” to answer 13.28.
9. How many grams of solute are required to saturate 100 g of water in each of the following solutions? (4 marks)
a. KCl at 80C
52 g
c. K2Cr2O7 at 50C
30 g
b. NaNO3 at 10C 80 g
d. KClO3 at 90C
46 g
10. What is each of the solutions below: saturated, unsaturated or supersaturated ? All of the solutes are mixed with
100 g of water. (4 marks)
a. 40 g of NaCl at 50C
supersaturated
b. 30 g of NaNO3 at 30C
unsaturated
c. 70 g of KCl at 20C
supersaturated
d. 86 g of Pb(NO3)2 at 50C
saturated
11. How many grams of KNO3 per 100 g of water would be crystallized from a saturated solution as the temperature
drops. Show your work. (8 marks)
a. 54C to 20C 100 g to 30 g = 70 g crystallized c. 54C to 40C100 g to 64 g = 36 g crystallized
b. 50C to 30C 88 g to 44 g = 44 g crystallized d. 40C to 0C 64 g to 14 g = 50 g crystallized
12. How many additional grams of K2Cr2O7 are required to keep each of the following K2Cr2O7 solutions saturated
during the temperature changes indicated ? Show your work.
(8 marks)
a. 100 g of water with a change of 10C to 30C ?
8 g to 16 g = 8 g must be added
b. 200 g of water with a change of 10C to 30C ?
2 x (8 g to 16 g) = 16 g must be added
c. 100 g of water with a change of 40C to 90C ?
22 g to 68 g = 46 g must be added
d. 1000 g of water with a change of 40C to 90C ? 10 x (22 g to 68 g) = 460 g must be added
e. 100 mL of water with a change of 10C to 60C ? 1 mL H2O = 1 g H2O
8 g to 38 g = 30 g must be added
13. At
a.
b.
c.
d.
what temperature are the following solutes equally soluble in 100 g of water ? (4 marks)
NaCl and KNO3
24C
K2Cr2O7 and Ce2(SO4)3
12C
KCl and KNO3
22C
O2 and CO
they are never equally soluble
14. Which solute is least affected by the temperature changes ? Give one answer for the salts and one for the gases.
(2 marks)
NaCl for the salts, He for the gases
15. Which salt decreases in solubility as temperature increases? (1 mark)
Ce2(SO4)3
16. How does the solubility of all “ionic solids” change with an increase in temperature ? Explain. (3 marks)
 as a rule, ionic solids solubility increases with an increase in temperature
 if solvation is an endothermic process, the more energy added, the greater the solvation
17.

How does the solubility of all “gases” (CH4, O2, CO and He) change with increased temperatures ? Explain.
(3 marks)
as a rule the solubility of gases decreases with an increase in temperature
gases are generally non-polar. As temperature rises the greater kinetic energy of the system forces them out
of solution as a result of their weak attraction for the solvent.
18. Complete 13.32
19. 19.2 g of sodium chlorate is dissolved in 689 g of water.
a) Calculate the mass percent of sodium chlorate in solution.
Mass % = Mass of solute
x 100 % =
Total Mass of Solution
19.2 g
19.2 g + 689 g
x 100 % = 2.71 %
b) Calculate the mass percent of sodium ions in solution.
NaClO3
19.2 g x
22.99 g/mol
(22.99 + 35.45 + 3(16.00)) g/mol
Mass % = Mass of solute
x 100 % =
Total Mass of Solution
= 4.14 g of compound is Na
4.14 g
19.2 g + 689 g
x 100 % = 0.586 %
20. Chromium is found in waste water at a mass percent of 0.00215 %.
a) What mass of chromium would be found in 500 000 L of waste ?
0.00215 %
100 %
x 500 000 L x
1 kg
1L
= 10.8 kg Cr
b) What is the concentration of chromium in ppm ?
1 ppm =
1 µg solute
1 g solution
=
1 mg
1 kg
=
10.8 kg x
500 000 kg
106 mg
1 kg
= 21.5 ppm
21. Sulfur dioxide is present in air at a concentration of 65 ppb, weight per weight of air.
a) What mass of sulfur dioxide is found in a classroom that has dimensions of 6.00 m by 8.00 m by 2.5 m ? The
density of air is 1.25 x 10-3 g/mL.
Mass of air = 1.25 x 10-3 g/mL x (1 mL / 1 cm3) x 6.00 m x 8.00 m x 2.5 m x ((102 cm)3 / (1 m)3)
= 1.48 x 105 g
65 ppb =
65 ng SO2
1 g air
x 1.48 x 105 g = 0.00959 g SO2
b) What would the mass percentage if the sulfur dioxide were dissolved in 500. mL of water?
0.00959 g SO2
500 mL x 1g/1mL
x 100 % = 0.00192 %
22. A commonly purchased disinfectant is a 3.0% (by mass) solution of hydrogen peroxide in water. Assuming that
the density of the solution is 1.0 g/cm3, calculate the molarity, molality and mole fraction of the H2O2.
3.0 %
100 %
x 1000 g solution = 30 g H2O2 in 1L of solution, given a 1.0 g/cm3 density
H2O2 = 34.02 g/mol; 30 g/34.02 g/mol = 0.88 mol/L H2O2
if the H2O2 mass = 30. g in 1 L of solution, then 1000 g of solution contains 1000 g – 30.g = 970 g H2O
= 0.97 kg
Molality =
0.88 mol H2O2 = 0.91 mol/kg
0.97 kg H2O
970 g H2O
18.02 g/mol
= 53.8 mol H2O
Mole Fraction =
0.88 mol H2O2
0.88 mol H2O2 + 53.8 mol H2O
= 0.016
23. The most concentrated aqueous solution of NaOH that can be prepared is approximately 50% NaOH by mass.
Calculate the mole fraction and molality of NaOH in this solution.
50 % NaOH: in 100g total mass, 50 g is NaOH and 50 g is H2O (1.25 mol NaOH, 2.77 mol H2O)
Mole Fraction =
Molality =
1.25 mol NaOH
1.25 mol NaOH + 2.77 mol H2O
1.25 mol NaOH
50 g H2O (1 kg/1000g)
= 0.311
= 25 mol/kg
24. A solution is prepared by dissolving 50.0 g cesium chloride in 50.0 g water. The density of the solution is 1.58
g/cm3. Calculate the mass percent, molarity, molality, and mole fraction of the cesium chloride.
Mass % =
50.0 g CsCl
x 100 % = 50.0 %
50.0 g CsCl + 50.0 g H2O
Volume of solution = 100.0 g
1.58 g/cm3
Moles CsCl =
= 63.3 mL (1 l / 1000 mL) = 0.0633 L
50.0 g
= 0.297 mol
(132.91 + 35.45) g/mol
Molarity CsCl =
Molality CsCl =
0.297 mol
0.0633 L
0.297 mol
0.0500 kg H2O
= 4.69 mol/L
= 5.94 mol/kg
Moles H2O =
50.0 g
= 2.77 mol
18.02 g/mol
Mole Fraction CsCl =
0.297 mol
= 0.0967
0.297 mol + 2.77 mol
25. A bottle of wine contains 12.5% ethanol by volume. The density of ethanol is 0.79 g/cm 3. Calculate the
concentration of ethanol in wine in terms of mass percent and molality.
Assume 1000. mL of wine:
12.5 % x 1000. mL = 125 mL ethanol, 1000. mL – 125 mL = 875 mL H2O
100 %
= 875 g
Mass of ethanol = (0.79 g/cm3)(125 mL) = 98.8 g
Moles of ethanol =
98.8 g
46.08 g/mol
Mass % C2H5OH =
98.8 g
x 100 = 10.1 %
98.8 g + 875 g
Molality C2H5OH =
2.14 mol
0.875 kg
= 2.14 mol
= 2.45 mol/kg
26. How would one prepare 3.5 L of 0.55 M NaCl from solid NaCl?
3.5 L x 0.55 mol/L x (22.99 + 35.45) g/mol = 110 g of NaCl is placed in a flask and diluted to a total volume of
3.5 L.
27. How would one prepare 3.00 x 102 g of 0.115 m ethylene glycol (C2H6O2) from ethylene glycol and water?
Total mass is 300. g
Molar mass of C2H6O2 = 62.08 g/mol
0.115 mol/kg = 1.15 x 10-4 mol/g =
(x g / 62.08 g/mol)
300. g – x g
x = 2.13 g ethylene glycol is added to (300. g – 2.13 g) = 297.87 g water
Complete 13.56 to 13.74, even