Assignment 7: Yet More Solubility Curves
20g
5g
20g
1. At 80oC, 100 grams of water is saturated with potassium chloride. How many
grams of solute will precipitate when the solution is cooled to 45oC
82g at 50oC, 62g at 40oC
82g – 62g = 20g ppt.
2. If 50 grams of water saturated with potassium chlorate at 23oC is slowly
evaporated to dryness, how many grams of salt will be recovered?
10gKClO3 = xg KClO3
x = 5g
100g H2O
50g H2O
3. If 30 grams of potassium chloride is dissolved in 100 grams of water, how
many additional grams would be needed to make the solution saturated at
80oC?
50g will go in 100g H2O at 80oC; 50g – 30g = 20g
51g
4. What is the smallest mass of water required to dissolve completely 23 grams of
ammonium chloride at 40oC?
45g NH4Cl = 23g NH4Cl
x = 51g H2O
100g H2O
x g H2O
1g/oC 5. A saturated solution of sodium nitrate in 100 grams of water at 40oC is heated
to 50oC. What is the rate of increase in solubility in grams per degree??
105g NaNO3 at 40oC
115g NaNO3 at 50oC
NH4Cl 6. Which chloride has the greatest percent by mass at 60oC? What is the percent?
55g NH4Cl x 100 = 35%
155g total
0.18M 7. What is the molarity (molality) of a saturated solution of cerium(III) sulfate at
25oC?
10g Ce2(SO4)3 x 1 mole Ce2(SO4)3 = 0.018 moles
568g Ce2(SO4)3
M = 0.018 moles Ce2(SO4)3 = 0.18M
0.100 L solution
0.020 8. What is the Ksp of a saturated solution of cerium(III) sulfate at 25oC?
Ce2(SO4)3 (s)   2 Ce+3 (aq) + 3 SO4-2 (aq)
[0.18]
0
0
- 0.18
+ 0.36
+ 0.54
0
[0.36]
[0.54]
Ksp = [Ce+3]2 [SO4-2]3 = [0.36]2 [0.54]3 = 0.020
-30 C 9. A solution of KNO3 is saturated at 50oC. At what temperature would the
solution freeze
82g KNO3 dissolve in 100g H2O at 50oC.
o
82g KNO3
x
1 mole KNO3 = 0.81 moles
101g KNO3
m = 0.81 moles = 8.1m
0.100 kg H2O
KNO3 (s)   K+1 (aq) + NO3-1 (aq)
[8.1]
0
0
-8.1
+8.1
+8.1
0
[8.1]
[8.1] = 16.2 m total particles
tf = 1.86oC/m x 16.2m = - 30oC
3.3m 10. A solution of KNO3 freezes at -12oC. At what temperature would the solution
be saturated?
-12oC = 1.86oC/m x m
m = 6.5m total particles
Because KNO3 breaks into 2 particles and the total is 6.5, it must be half that for
each ion or 3.3m each
KNO3 (s)   K+1 (aq) + NO3-1 (aq)
[3.3]
0
0
-3.3
+3.3
+3.3
0
[3.3]
[3.3]
= 6.5m total
Assignment 8: Molarity, molality, and colligative properties
1. What is the molarity of a solution in which 82.0g of calcium nitrate is dissolved in
enough water to make 500.0 mL of solution?
82.0g Ca(NO3)2 x 1 mole = 0.4998 moles
164.09g
M = 0.4998 mol = 0.100M
0.5000 L soln
2. What is the molality of a solution in which 50.0g of copper(II) sulfate is dissolved
in 250.0mL of water
50.0g CuSO4 x 1 mole CuSO4 = 0.313 moles
159.5 g CuSO4
m = 0.313 moles CuSO4 = 5.02 m
0.2500 kg H2O
3. Calculate the mass of solute in 250.0mL of sodium sulfate solution that is 2.00M
(molar).
2.00M = x moles Na2SO4 = 0.500 moles
0.2500 L soln
0.500 moles Na2SO4 x
122.04g Na2SO4 = 61.0g Na2SO4
1 moles Na2SO4
4. Calculate the mass of solute in 250.0mL of sodium sulfate solution that is 2.00m
(molal).
2.00M = x moles Na2SO4 = 0.500 moles
~ 0.2500 kg H2O
0.500 moles Na2SO4 x
122.04g Na2SO4 = 61.0g Na2SO4
1 moles Na2SO4
5. The Ksp of silver iodide is 8.3 x 10-17 at 25oC; what is the molarity of a saturated
solution? Write the reaction for the dissolving.
AgI (s)   Ag+1 (aq) +
x
0
0
-x
+x
+x
0
x
x
Ksp = [Ag+1] [I-1]
8.3 x 10-17 = x2
x = 9.1 x 10-9 M
I-1 (aq)
6. A saturated solution of lead(II) chloride at 25oC contains 2.2 grams of solute in
500.0 mL. What is the Ksp? Write the reaction for the dissolving.
2.2g PbCl2 x
1 mole PbCl2
278.1g PbCl2
= 7.9 x 10-3 moles
M = 7.9 x 10-3 moles = 0.016 M
0.5000 L
PbCl2 (s)   Pb+2 (aq)
[0.016]
0
- 0.016
+ 0.016
0
[0.016]
Ksp = [Pb+2] [Cl-1]2
+ 2 Cl-1 (aq)
0
+ 0.032
[0.032]
= [0.016] [0.032]2 = 1.6 x 10-5
Is the lead(II) chloride or the silver iodide in the previous problem more soluble?
17
The lead(II) chloride is more soluble as the Ksp is higher (1.6 x 10-5 vs 8.3 x 10)
7. What is the freezing point of a 0.85 molal solution of sugar? Kf = 1.86oC/m.
C12H22O11 (s)   C12H22O11 (aq)
0.85
0
- 0.85
+ 0.85
0
0.85
tf = 1.86oC/m x 0.85m = 1.6oC
so the freezing pt. is – 1.6oC
8. What is the freezing point of a solution that contains 68.5 grams of sucrose,
C12H22O11, dissolved in 100.0 grams of water?
68.5g C12H22O11 x
1 mole
342g
= 0.200 moles
m = 0.200 moles = 2.00 m
0.1000 kg
C12H22O11 (s)   C12H22O11 (aq)
2.00
0
- 2.00
+ 2.00
0
2.00
tf = 1.86oC/m x 2.00m = 3.72oC
so the freezing pt. is – 3.72oC
9. What is the freezing point of a solution that contains 68.5 grams of salt, NaCl,
dissolved in 100.0 grams of water?
68.5g NaCl x 1 mole = 1.17 moles
58.5g
m = 1.17 moles = 11.7 m
0.1000 kg
NaCl (s)   Na+1 (aq) + Cl-1 (aq)
11.7
0
0
- 11.7
+ 11.7
+11.7
0
11.7m
11.7m
23.5m total
tf = 1.86oC/m x 23.4m = 43.5oC
so the freezing pt. is – 43.5oC
Why is CaCl2 used on the roads instead of NaCl or sucrose?
When CaCl2 dissolves, 3 particles are produces. When NaCl dissolves, only 2
particles are produced.
10. What is the boiling point of the solution in problem #9?
68.5g NaCl x 1 mole
58.5g
= 1.17 moles
m = 1.17 moles = 11.7 m
0.1000 kg
NaCl (s)   Na+1 (aq) + Cl-1 (aq)
11.
0
0
- 11.7
+ 11.7
+11.7
0
11.7m
11.7m
23.5m total
tb = 0.512oC/m x 23.5m
tb = 12oC
so the boiling pt. is 112oC