Chem 1A
1 Dr. White
Workbook 3 – Problems for Exam 3
3-1: Types of Solids
1. What type of crystal will each of the following substances form in its solid state? Choices to consider are
ionic, metallic, network, noble gas, nonpolar molecular, polar molecular.
(a) C2H6 __________
(b) Na2O ____________
(c) SiO2 ______________
(d) CO2 ______________
(e) NO __________
(f) NaNO3 ______________
(g) Al ________________
(h) C(diamond) __________ (I) SO2 ________________
2. Circle all the compounds in the following list which would be expected to form intermolecular hydrogen
bonds in the molecular solid state:
(a) CH3OCH3
(b) CH4
(c) HF
(d) CH3COOH
(e) Br2
(f) CH3OH
(dimethyl ether)
(acetic acid)
(methanol)
3. Specify the predominant Intermolecular Force of Attraction (Dispersion, Dipole-Dipole, or H-Bonding) or
Bonding Force (ionic, metallic, or network covalent) involved for each solid in the space immediately following
the substance. Then in the last column, indicate which member of the pair you would expect to have the higher
melting point.
Solid #1*
Predominant Force
Solid #2
Predominant Force Substance with Higher
Melting Point
(a) HCl
I2
(b) CH3F
CH3OH
(c) H2O
H 2S
(d) SiO2
SO2
(e) Fe
Kr
(f) CH3OH
CuO
(g) NH3
CH4
(h) HCl
NaCl
(i) Diamond
Cu
*Note: Many of the compounds given are not solids at room temperature. But if you cool them down to a low
enough temperature, eventually they will become solids.
3-2: IMFs
1. List the dominant IMFA in each of the following substances:
Pentane (C5H12)
Methanol (CH3OH)
water
ammonia
hydrochloric acid
carbon tetrachloride
CHCl3 (chloroform)
2. If the two substances listed below in each case were interacting, what IMFA would dominate the interaction?
NH3 and HF
XeF4 and Br2
Hexane (C6H14) and pentane (C5H12)
HCl and H2O
3. Circle all of the species below that can form a hydrogen bond. Explain why the other species couldn't
hydrogen bond.
C 2H 6
CH3NH2
HCl
CH3CH2CH2OH
4. Which of the following species should have the weakest intermolecular forces? Why? (rank from weakest
to strongest)
Cl2
I2
N2
H 2O
5. Which of the following materials should have the highest boiling point? Why?
highest to lowest)
H2Se
H 2S
H2Po
H2Te
6. Which of the following materials should be the most viscous? Why?
HOCH2CH2OH
Br2
CH3(CH2)3CH3
(rank the boiling points from
Chem 1A
2 Dr. White
7. Which of the following substances should have the highest melting point? Why?
I2
H 2O
F2
NaI
8. Determine the species that you would expect to have the desired property and briefly explain your choice.
More viscous
Property
Species
CH3CH2CH2CH2OH or CHCl3
Higher Boiling Pt
Br2 or H2Se
More likely to liquefy (or condense)
Br2 or F2
Greater Capillary action in a glass (SiO2) tube
H2O or Hg
3-3: Solids: Unit Cells and Semicomductors
1. Platinum, Pt, crystallizes in the unit cell structure shown to the left.
a. Identify this unit cell type.
b. How many net atoms are contained in one unit cell of this type?
c. What would the edge length be for this unit cell type in terms of r?
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d. If Pt has a density of 21.45 g/cm , what is the edge length of the unit cell in
pm?
2. The metallic radius of manganese is 127 pm. If it crystallizes in a body3
centered cubic unit cell, what is its density in g/cm ?
3
3. The density of sodium metal is 0.971 g/cm and the unit cell length is 428.5 pm. Determine the unit cell of
sodium metal. (Hint: Determine the # of atoms/unit cell)
3
4. Iron adopts a BCC unit cell with an atomic radius of 0.124 nm. What is its density in g/cm ?
3
5. Iridium adopts a FCC unit cell and a density of 22.56 g/cm . What is the radius of an Ir atom in nm?
6. Shown below is single unit cell
To which cubic structure do the calcium ions adopt?
Also, how many atoms of each are contained in a unit
cell. What is the simplest whole number ratio?
-
7. Calculate the edge length of the unit cell of NaCl on the basis on of the average radii of Cl
+
+
(181 pm) and Na (98 pm) assuming that the corner Cl ions and the Na along the edges touch
(see figure to the right).
8. Lithium chloride crystallizes with the same structure as NaCl except that the chloride anions touch along the
face diagonal of the unit cell. If the unit cell edge is 513 pm, what is the radius of a chloride ion? (hint – use the
Pythagorean theorem…what is the face diagonal equal to given the first sentence of this problem?)
9. Iron may crystallize in either the body-centered cubic or face-centered cubic structures. Calculate the
densities of iron in these structures, given that the radius of an iron atom is 124 pm.
Chem 1A
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10. You are given a small bar of an unknown metal. You find the density of the metal to be 10.5 g/cm . An X-10
ray diffraction experiment measures the edge of the face-centered cubic unit cell as 4.09 Å. (1 Å = 1 x 10 m).
Identify this metal.
11. Draw band diagrams for conductors, semiconductors, and insulators and explain how the band gap
influences the conductivity using these diagrams.
12. Define p-type and n-type semiconductors, and list what elements would be appropriate to add to Si to make
each type of extrinsic semiconductor.
13. Describe how the band structure of Si is affected by addition of elements to make p- and n-type
semiconductors. Draw pictures of band structures to illustrate these changes.
14. In each of the following, determine which has the highest melting point. Explain your answer.
a. Rb (s) or Sr (s)
b. Zn (s) or Cd (s)
c. SrCl2 (s) or RbCl (s)
d. MgBr2 (s) or MgI2 (s)
3.4 Heating Curves and Phase Diagrams
1. Liquid ammonia (boiling point = -33.4°C) can be used as a refrigerant and heat transfer fluid. How much
energy, in kJ, is needed to heat 25.0 g of NH3(l) from -65.0°C to -12.0°C? ?
NOTE: J/(g•K) is the same as J/(g•°C)
2. Diethyl ether, used as a solvent for extraction of organic compounds from aqueous solutions, has a high
vapor pressure which makes it a potential fire hazard in laboratories in which it is used. How much energy is
released when 100.0 g is cooled from 53.0°C to 10.0°C?
boiling point
34.5°C
heat of vaporization
351 J/g
specific heat capacity, (CH3)2O(l)
3.74 J/(g·K)
specific heat capacity, (CH3)2O(g)
2.35 J/(g·K)
Pressure in torr
3. Use the phase diagram below to answer the questions below.
a) What two phases are in equilibrium at the
circle?
2500
b) Consider the point represented by the
Liquid
triangle.
Solid
2000
At the Pressure of the triangle (keeping it
▲
constant), what is the temperature required
1500
to convert this substance into a solid?
f)
1000
500
Gas
At the Temperature of the triangle (keeping
it constant), what is the pressure required
to convert this substance into a gas?
c) What is the approximate normal melting point
of this substance? Is there one? Why or why
not?
10
20
30
40
50
60
70
d) What is the approximate boiling point of this
Temperature in 0C
substance at a pressure of 2 atmospheres?
e) Identify the triple point and the critical point of
this substance (write in on graph).
Under standard thermodynamic conditions (1 atm, 25˚C), what is the physical state of this material?
Chem 1A
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3-5 Properties of Solutions
1. Determine if each species listed below would be a strong electrolyte, weak electrolyte, or a
nonelectrolyte.
a. Glucose (C6H12O6)
b. KCl
c. HCl
d. HNO2
2. How many moles of lithium chlorate are in 153 mL of a 1.764 M solution?
3. Determine the mass (g) of solute required to form 250.0 mL of a 0.250 M sodium cyanide.
4. A 25.0 mL sample of concentrated HCl (12.0 M) is diluted to a final volume of 750.0 mL. What is the
molarity of the final solution?
5. A solution is prepared by dissolving 516.5 mg of oxalic acid (C2H2O4) to make 100.0 mL of solution. A
10.00 mL portion is then diluted to 250.0 mL. What is the molarity of the final solution?
6. An experiment calls for the use of 25.0 mL of a 0.100 M glucose solution. All you have available is a
2.00 M stock solution of glucose. How would you prepare the desired solution?
3-6 Concentration Units
1. An aqueous solution of H3PO4 is 25.0% by mass and has a density of 1.39 g/mL. Fill out the following table.
Solute =
Solvent =
Solute
Relative Mass (g)
# Moles
Volume (ml)
----
Solvent
----
Solution (total)
Now, calculate the following concentrations:
(a) Solution Molarity (M).
(b) Solution Normality (N).
(c) Solution molality (m).
(d) Mole fraction of H3PO4.
2. A 0.750 M solution of H2SO4 in water has a density of 1.046 g/mL at 20°C. Fill out the table below.
Solute =
Solvent =
Solute
Mass (g)
# Moles
Volume (ml)
----
Solvent
----
Solution (total)
Now, calculate the following concentrations:
(a) Solution Normality (N).
(b) Solution molality (m).
(c) Mole fraction of H2SO4.
3. An aqueous 1.06 m solution of CaCl2 (MM = 110.9 g/mol) has a density of 1.07 g/mL. Determine the molarity
and mass percent CaCl2. (feel free to use a table!)
4. Suppose you want to make another batch of CaCl2 solution with the same molality (m) as in the problem
above. You only have 7.00 g of CaCl2 (s) available. What mass of water (in grams) should you use?
3-7 Solubility
1. Predict whether each of the following is more likely to dissolve in CCl4 or water.
a. C7H16
b. Na2SO4
c. HCl
d. I2
2. Which of the following in each pair is likely to be the more soluble in water:
Chem 1A
Dr. White
5 3. Which of the following in each pair is likely to be more soluble in CCl4:
5. State the dominant IMF between each solute and solvent pair given below, and then indicate whether or not
you would predict the solute to dissolve well in the solvent.
Will solute
Solvent
Dominant IMF between solute & solvent
dissolve well
Solute in solvent?
NH3 (l)
H2O (l)
Br2 (l)
H2O (l)
C9H19OH (l)
C6H14 (l)
b) Would you expect ammonia to dissolve in water? Why or why not?
c) Would you expect iodine to dissolve in water? Why or why not?
d) What would be a good solvent in which you could dissolve I2? Explain.
3-8 Vapor Pressure and Colligative Properties
1. The vapor pressure of acetone (CH3COCH3) at 25 °C is 271 torr and that of methanol is 143 torr. What is the
vapor pressure of a solution that is 27.5 % methanol by mass in acetone as the solvent?
2. A solution has a 1:4 ratio of pentane to hexane (this means that if there are 5 total moles, 1 mole would be
pentane and 4 would be hexane). The vapor pressures of the pure hydrocarbons at 20°C are 441mmHg for
pentane and 121mmHg for hexane. What are the partial pressures of the 2 hydrocarbons above the solution?
What is the total pressure above the solution?
3. Calculate the vapor pressure of a solution prepared by dissolving 175 g of glucose (MM = 180.16 g/mol) into
350.0 mL water at 75°C. The vapor pressure of pure water st 75°C is 289.1 mm Hg and its density is 0.97489
g/mL.
4. If equal amounts of each of the following nonvolatile solutes is added to water to make a solution, which
solute will result in larger changes in vapor pressure, boiling point and freezing point? Why? (You should be
able to answer this question without doing any math)
i)
Glucose (C6H12O6)
ii)
NaCl
iii)
MgCl2
iv)
K3PO4
5. 0.134 moles of an unknown solute is dissolved in 50.0 g of water to make a solution with a boiling point of
104.1˚C. Which of the following solutes was used to make this solution? Fructose(C6H12O6), sodium bromide,
calcium carbonate, potassium sulfate or aluminum chloride. Show all our work and briefly explain your choice.
6. A physician studying a type of hemoglobin formed during a fatal disease dissolves 21.5 mg of the protein in
˚
water at 5.0 C to make 1.5 ml of solution in order to measure its osmotic pressure. At equilibrium, the solution
has an osmotic pressure of 3.61 torr. What is the molar mass (g/mol) of the hemoglobin?
7. Calculate the molality and experimental van’t Hoff factor i for an aqueous solution with 1.00 mass % NaCl,
freezing point = -0.593˚C.
Chem 1A
6 Dr. White
8. Use the following curves to answer the following questions: Vapor Pressure Curves for Water & an Aq. Solution of
Glycerol!
280!
300!
320!
340!
200!
180!
Vapor Pressure (mm Hg)!
160!
140!
120!
100!
80!
60!
40!
20!
0!
Temperature (K)!
a. Which curve corresponds to pure water?
b. Which curve corresponds to the solution?
c. What is the vapor pressure lowering ΔP at 320 K ?
d. What is the mole fraction of solute in this solution at 320 K ? (What is vapor pressure of pure water at
320 K ?)
e. What is the boiling point elevation ΔTb at 130 mm Hg? Label this on the graph.
f. If Kb for water is 0.512˚C/m, what is the molality of this solution?
g. How much glycerol (C3H8O3) in grams must be added to 1.00 kg of water to elevate the boiling point
by 10.0 ˚C?
6. In the phase diagram below, label the regions corresponding to the solid, liquid and gas phases of this
particular solvent. Then draw in how this diagram would change upon addition of a nonvolatile solute to this
solvent, labeling all the new curves you draw in and any important quantities, like ΔP, ΔTbp, ΔTfp.
Pressure (atm) Temperature Chem 1A
7 Dr. White
3-1: Types of Solids
1. (a) C2H6 nonpolar
(b) Na2O ionic
(c) SiO2 network
(d) CO2 nonpolar (e) N2O5 polar
(f) NaNO3 ionic
(g) Al metallic
(h) C(diamond) network
(I) SO2 polar
2.
(a) CH3OCH3
(b) CH4
(c) HF
(d) CH3COOH
(e) Br2
(f) CH3OH
(dimethyl ether)
(acetic acid)
(methanol)
3.
Solid #1*
Predominant Force
Solid #2
Predominant Force Substance with Higher
Melting Point
(a) HCl
DDA
I2
LDF
HCl
(b) CH3F
DDA
CH3OH
H-bonding
CH3OH
(c) H2O
H-bonding
H 2S
DDA
H 2O
(d) SiO2
network covalent
SO2
DDA
SiO2
(e) Fe
Metallic
Kr
LDF
Fe
(f) CH3OH
H-bonding
CuO
Ionic
CuO
(g) NH3
H-bonding
CH4
LDF
NH3
(h) HCl
DDA
NaCl
Ionic
NaCl
(i) Diamond
network covalent
Metallic
Diamond
Cu
3-2: IMFs
1. Pentane - LDFs
Methanol – H-bonds
water H-bonds
hydrochloric acid – DDAs
carbon tetrachloride - LDFs
ammonia – H-bonds
CHCl3 - DDAs
2. NH3 and HF
H-bonds
HCl and H2O
DDAs
XeF4 and Br2
LDFs
Hexane and pentane
LDFs
3. C2H6
CH3NH2
HCl
CH3CH2CH2OH
C2H6 and HCl do not H-bond because there are no H’s bonded to N, O, or F.
4. N2 < Cl2 < I2 < H2O
The weakest IMFs are in N2 because it is nonpolar and only has LDFs. The LDFs are
weaker than that in the other nonpolar molecules because it is the smallest and has the least number of
electrons.
5. Highest b.p. will be H2Po, then H2Te, then H2Se, then H2S. All have dipole-dipole interactions. So, they are
ranked from largest to smallest size since the largest atoms have the most electrons and thus are the most
polarizable and will have stronger LDFs.
6. HOCH2CH2OH – because it has the strongest IMFs (H-bonds).
7. NaI because it has the strongest forces holding the solid together. It has ionic bonds while the rest have
IMFs.
8.
Property
More viscous
Higher Boiling Pt
More likely to liquefy (or condense)
Greater Capillary action in a glass (SiO2) tube
Species
CH3CH2CH2CH2OH or CHCl3 It has H-bonds and CHCl3
has weaker DDAs.
Br2 or H2Se It has DDAs and Br2 only has weaker LDFs
Br2 or F2 It is larger and forms stronger LDFs than F2.
This means that it will want to stay a liquid and not go to
the gas phase
H2O or Hg Water forms strong interactions with the glass
tube, Hg does not.
Chem 1A
8 Dr. White
3-3 Solids, Unit Cells, and Semiconductors
1. a. FCC
b. 4.00 atoms
c.
3
2. 7.23 g/cm
3. BCC
3
4. 7.90 g/cm
5. 0.136 nm
2+
6. FCC; 4 Ca :8F ; CaF2
€
7. 558 pm
8. 181 pm
3
3
9. FCC: 8.60 g/cm and BCC: 7.90 g/cm
10. MM = 108.2 g/mol, therefore Ag
11.
2 2r
d. 392 pm
A semiconductor is on the left, the insulator is
in the middle and the conductor is on the right.
The size of the band gap dictates the
conductivity. The smaller the band gap, the
more conductive the material.
12. N-type semiconductors are made by adding atoms with more valence electrons to an intrinsic
semiconductor. A “donor level” is created between the valance and conductance bands. Therefore the band
gap is decreases and conductivity increases. We can add P (or any element with more valence electrons) to Si
to make an n-type semiconductor. P-type semiconductors are made by adding atoms with less valence
electrons. Positive “holes” are created in the valence band. The electrons can migrate to these sites and the
conductivity increases. We can add Al (or any other element with less valance electrons) to Si to make a p-type
semiconductor.
13.
e- e-
⊕
14.
⊕
e-
See description in #12. The figure on the left is a
p-type and the figure on the right is the n-type.
⊕
a. Sr has the higher melting point. Sr has a stronger metallic bond since it has more valence electrons
which creates stronger metallic bonds.
b. Zn has the higher melting point because it is smaller than Cd. This makes the bonds stronger and
thus the metallic bond is stronger. This translates to a higher melting point.
c. SrCl2 has a higher melting point because it has stronger ionic bonds due to the higher charge of the
strontium ion.
d. MgBr2 has a higher melting point because it has stronger ionic bonds due to the smaller size of the
bromide ion. This allows the ions to get closer and form stronger ionic bonds.
3-4 Heating Curves and Phase Diagrams
1. 39.4 kJ
2. -48.6 kJ
3. a. solid and gas
b. ~32°C and ~1300 torr
Chem 1A
9 Dr. White
c. There is no normal malting point. The solid goes directly to a gas at this point.
d. ~56°C
e. Triple point is where all three lines intersect and Critical point it the highest P & T on the liquid-gas line.
f. gas
3-5 Properties of Solutions
1. a. Nonelectrolyte
b. Strong Electrolyte
c. Strong Electrolyte
d. Weak electrolyte
2. 0.270 mol
3. 3.06 g
4. 0.400 M
-3
5. 2.295 x 10 M
6. Mix 1.25 mL of the 2.00 M glucose solution with enough water to make 25.0 mL total.
3-6 Concentration Units
1. 3.55 M, 10,6 N, 3.40 m, 0.0577
2. 1.50 N, 0.771 m, 0.0137
3. 1.01 M, 10.5%
4. 59.5 g H2O
3-7 Solubility
1. a. CCl4 b. water
c. water
d. CCl4
2. B, B, B, B, A
3. A, B, A
4. a.
NH3 (g)
H2O (l)
H bonding
yes
Br2 (l)
H2O (l)
LDFs
No
C9H19OH (l) C6H14 (l)
LDFs
Yes
b. I would expect ammonia (NH3) to dissolve in water, as the interaction between ammonia and water (H
bonding) is the same as the IMFs holding both water and ammonia together. Thus breaking apart both the
solute and solvent is compensated for by the strong interaction between solute and solvent.
c. Here I would not expect iodine to dissolve in water. Water has H-bonding and the nonpolar I2 only has LDFs
the new interaction between I2 and water is not stronger than H bonding in water. So water will not want to mix
with I2.
d. A nonpolar solvent, like hexane, would be a good solvent for I2 – like dissolves like.
3-8 Vapor Pressure and Colligative Properties
1.219 torr
2. pentane: 88.2 mmHg
hexane: 96.8 mmHg
Total: 185.0 mmHg
3.275 mmHg
4. The K3PO4 would result in the largest changes in vapor pressure, boiling point and freezing point. The
colligative properties all depend on the amount of solute present in solution. When dissolved in solution, the
K3PO4 dissociates into the most particles per mole of compound, 4 particles. Another way to say this is that i =
4 for K3PO4.
5. since i = 3, Only potassium sulfate, K2SO4, would be expected to break into 3 ions and thus have i = 3.
4
6. 6.9 x 10 g/mol
7. 0.174 m; i = 1.83
a. The solid curve belongs to pure water, it
8.
shows higher vapor pressure at any
temperature.
b. The dashed curve belongs to the solution,
it shows lower vapor pressure at any
temperature.
c.
Just estimate the difference between
the two curves at 320 K, this looks like
~12 mm Hg
d. From curve, vapor pressure of pure water
at 320 K is ~80 mm Hg, so χ = 0.15
e. Just estimate the difference between
the two curves at 130 mm Hg, this
looks like ~2 K = 2˚C to me
f. 4 m
3
g. 1.8 x 10 g glycerol
h.
Chem 1A
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