Key for: Worksheet for Chapters 15 and 16
Spring 2013
**For Problems 1-6, suggested pages to read in the book are: 445-458**
1.
-
-
Water has a high surface tension and a low vapor pressure. Illustrate/draw the property of surface
tension and vapor pressure from the perspective of the H2O molecules. [Hint: polar covalent
molecules]
Surface tension: The water molecules within the body of he liquid form hydrogen bonds with other
molecules that surround them on all sides. The attractive forces on each of the molecules are
‘balanced’. However, water molecules at the surface of a liquid experience an unbalanced attraction.
Water molecules at the surface of a water drop cannot form hydrogen bonds with air molecules, so
they are drawn into the body of the liquid and hydrogen bond with other water molecules within the
water droplet, producing surface tension.
Vapor pressure: Hydrogen bonding between water molecules also explains water’s unusually low
vapor pressure. Please remember that vapor pressure of a liquid is the result of molecules escaping
from the surface of the liquid and entering the gas/vapor phase. Because hydrogen bonding is the
strongest of the intermolecular forces, the tendency of individual water molecules to escape is low and
evaporation is slow (as compared to other substances).
2.
Sketch a picture of how an electrolyte behaves in water, the universal solvent.
- First remember that and electrolyte is a compound that conducts an electric current when it is in an
aqueous solution
-
3.
-
4.
-
Determine and explain the primary difference in behavior for a non-electrolyte when compared to
the behavior of an electrolyte.
Ionic compounds are electrolytes because they dissociate into ions and conduct electricity. A
compound that does not conduct an electric current is a nonelectrolyte. Nonpolar covalent
compounds are nonelectrolytes because they do not dissociate into ions when placed in water.
The equation for solid NaCl when placed in water is NaCl (s)  Na1+(aq)
What is the equation for solid Ammonium sulfate when place in water?
+ Cl1-(aq)
(NH4)2SO4(s)  2NH4+(aq) + SO42=(aq)
5.
Predict the mass, using calculations (stoichiometry) of the remaining product, when 100.0 g of the
hydrate, Magnesium sulfate heptahydrate is heated.
MgSO4 7H2O(s)

MgSO4 (s)
+
7 H2O (g)
100.0 g MgSO4  7H2O x 1 mol MgSO4  7H2O x 1 mol MgSO4 x
120.38 g MgSO4 = 48.83 g MgSO4
246.52 g MgSO4  7H2O 1 mol MgSO4  7H2O 1 mol MgSO4
6. If water is the universal solvent and is a polar covalent molecule predict, which substance below would
dissolve or dissociate.
-
Also classify the substance as ionic, covalent polar or covalent non-polar.
a. CH4 - Does not dissolve/dissociate. Covalent non-polar
b. KCl - Does dissociate. Ionic
c. He – Does not dissociate. It’s an atom that is completely neutral and is also nonpolar.
d. MgSO4 – Does dissociate. Ionic
e. CO2 - Does not dissociate. Covalent non-polar
** For Problems 7-11, suggested pages to read are: 459-462 & 470-491**
7. Using the table provided in the textbook (p. 462)
a. Draw or explain the Tyndall effect.
-
When particles in a solution are big enough in solution to scatter visible light. Particles in a solution are
too small to scatter the light, but are bigger in colloids and suspensions so you can see a beam of light
when it is passed through. This is one way to distinguish between a solution versus a colloid or
suspension.
b. Differentiate between the behavior of solutions verses colloids in respect to the Tyndall effect.
-
See answer to part a.
c. Name two ways to distinguish a suspension from a colloid.
-
First way is to let a suspension and a colloid rest for a while. The suspension will settle out when left
standing. The larger size of the suspended particles means that gravity plays a larger role in causing
them to separate out from the mixture.
-
A second way to distinguish between suspensions from a colloid is to filter the mixtures. The colloid
mixture has particles that are smaller than those in suspensions and thus cannot be retained by filter
paper where as the larger particles in a suspension can be separated when using filter paper.
8. Construct a graph for the most common solubility pattern for solid crystals. Place the solubility on the yaxis (dependent) and the temperature on the x-axis (independent).
9.
Using Henry’s Law: S1 =
S2
on p. 476 in the textbook;
P1
P2
Determine the solubility of CO2 gas at 2.0 atmospheres of pressure; if it has a solubility of 0.169 g/100 g H2O at
1.00 atmospheres of pressure.
-
S1
= .169g
S1 = .39 g CO2
2.0 atm
1.00 atm
100 g H2O
10.
Using the formula on p. 483: M1 V1 = M2 V2 ; calculate and explain how to make 250 mL of 1.00
M solution of HCl from concentrated HCl (12.0 M). You may use diagrams.
11.
(12.0 M) x (V1) = (1.00 M) x (250 mL)
V1 = 20.8 mL HCl
Distinguish between a 1 M (Molar) solution and a 1 m (molal) solution.
-
1 Molar = 1 mol of solute
1 liter of solution
- 1 molal = 1 kg solute__
1 kg of solvent
**For problems 12-18, suggested pages to read in the book are: 488-496**
12.
When we made ice cream we used the principle of freezing point depression. From the particle
interaction perspective draw or explain why adding NaCl to the ice/water will lower the freezing point.
-
When something freezes in the presence of a solute, it disrupts the formation of the orderly pattern
than occurs when a substance (such as water) freezes. As a result of the presence of a solute, more
Kinetic Energy must be used to move the solute particles out of the way in order for the solution to
solidify. Because it requires K.E to move the ions out of the way, this becomes P.E. and because there
is less K.E. available the temperature of the freezing point is lower. Remember that the K.E. is a
measured by temperature.
13.
Calculate the molality (m) of a solution prepared by dissociating 10.0 g NaCl in 600 g of H2O. [Hint:
water is the solvent and m = mol of solute
kg of solvent
-
10.0 g NaCl x 1 mol NaCl x 1000 g =
600 g H2O
58.44 g NaCl
1 kg
2. 85 m
 T = mol solute  Kf
 T = mol solute  Kb
kg solvent
kg solvent
There is a chart on p. 494 in the textbook for some Kf (molal freezing-point depression constants) values; using
the chart calculate the new freezing point of water, when 100.0 g of glucose (C6H12O6) is added to 250.0 g of
H2O (water is the solvent).
14.
Formulas:
-
100.0 g C6H12O6
-
250 g H2O x 1 kg
1000 g
-
 T = .5550 mol C6H12O6 x (1.86 °C kg) =
.250 kg H2O
mol
x 1 mol C6H12O6
180.18 g C6H12O6
= .5550 mol C6H12O6
= .250 kg H2O
4.13 °C
15.
If 99.0 grams of non-ionizing (non-electrolyte) solute are dissolved in 669 grams of water, and the
freezing point of the resulting solution is 0.960oC, what is the molecular mass of the solute?
-
 T = mol solute  Kf
kg solvent
.960 °C = mol solute x (1.86 °C kg)
.669 kg H2O
mol
mol solute = .345 mol
-
Molar mass = grams
mol
16.
99.0 grams = 287 g
.345 mol
mol
Look at the graph for solubility vs. temperature for gases, shown below.
a.) Interpret the trend the graph illustrates.
-
The solubility of gases decreases as the temperature increases
b.) What is the solubility of CO2 gas at 20.0oC compared to the solubility of CO2 gas at 35.0oC?
-
At 20°C = ~ 1.8 g CO2
100 g H2O
- At 35°C = ~ 1.4 g CO2
100 g H2O
c.) Speculate on how an increase in the temperature of the water in a stream or lake would have
a negative impact on the environment.
-
Less O2 available for creatures and or plant life in the lake/stream that depend on oxygen gas to
survive.
17.
If a certain amount of solute is added to 25.0 g of benzene and that solution is found to freeze at a
temperature of 1.96oC below the normal freezing point of pure benzene, what number of moles (mol) of
solute was added to the solution?
5.12
The Kf of benzene is
-
C kg
mol
T  mk f
1.96 °C = mol solute x (5.12 °C kg)
.0250 kg benzene
mol 


m
mol
kg
mol solute = .00957 mol or 9.57 x 10-3 mol
18.
Considering the above moles of solute determined, what is the molar mass of the solute if 1.42 grams
of solute was added to the solution to cause the determined change in freezing point?
-
Molar mass =
1.42 g solute
9.57 x 10-3 mol solute
= 148 g
mol
**For questions 19-25 review portions of chapter 15 and 16 to answer them**
Using the list of solutes below, answer questions 19-25. (You should list all the correct answers for each
question).
a. C4H10 (l)
b. NaCl (s)
c. AlCl3 (s)
d. CuCl2 (s)
e. CH4 (g)
19.
Which would cause the greatest boiling point elevation? Answer = C. AlCl3 (s)
20.
Which would be an electrolyte? Answer = b. NaCl(s), c. AlCl3(s), and d. CuCl2(s)
21.
Which would be less soluble in H2O with an increase in temperature? Answer = e. CH4(g)
22.
Which would be a non-electrolyte? Answer = a. C4H10 (l) and e. CH4(g)
23.
Which is volatile? Answer = a. C4H10 (l)
24.
Which would be a solution with color? Answer = d. CuCl2
25.
Which would cause the greatest freezing point depression? Answer = C. AlCl3 (s)
26.
Explain one of the above answers with a descriptive labeled diagram
-
Student answers will vary, but need if labeling and drawing the boiling point elevation or freezing point
depression, it should look like the figure 16.20 on page 494 in the textbook.