WA CP Chemistry Exam 4 MC review Set B 1. A student is attempting to dissolve two different alcohols, Ethanol and Butanol, in
water and tries to predict which one will dissolve more thoroughly.
a. Ethanol because its molecules are smaller.
b. Ethanol because it has a higher percentage of polar area.
c. Butanol because it has more hydrogen bonding.
d. Neither, as both have a single hydroxyl group, and are thus equally polar.
2.Identical containers A and B holding samples of liquid H2O are heated at a constant rate
from 0° C. Container A has 10.0g of water and an interior vapor pressure of 1.0 atm,
while B has 20.0g of water and a vapor pressure of 1.0 atm. In which container will the
water boil first?
a. Container A, as it has less water and therefore a lower heat of vaporization than B
b. Container A, as heat of vaporization depends on mass, and A has less mass, requiring
less energy to completely boil the sample
c. Container B, as a greater surface area of the water would be heated due to a greater
volume
d. Neither, since the boiling points would be identical for both samples and the heating
is at a constant rate
3. Data for the following question may not be accurate to real Krypton gas. A sample of
Krypton gas is placed inside of a 1.0 L container at a temperature of 12.0° C, and exerts a
pressure of 350. torr. If heated to 24.0° C, what will its new pressure be?
a. 0.48 atm
b. 365 torr
c. Both A and B
d. 700. Torr
4. Which of the following experiences the greatest Intermolecular Forces?
(A) Formaldehyde
(B) Ammonia
(C) Un-Iodized Table Salt
(D) Ethyl Alcohol
Refer to the table below for questions 5-6
Moles
N2
Pressure
1.2 mol
H2
4 atm
I2
1.4 mol
Total :
4.2 mol
A student collected the preceding data, base questions 2-3 on that table. All
three gases are placed into the same container, at the same temperature.
___ 5. What is the partial pressure of the I2 in the mixture?
(A) 3.5 torr
(B) 3.3 atm (C) 1660 torr
(D) 2660 mm Hg
___ 6. If the mixture of gas were to be moved into a 5 liter sealed and rigid
container, what would the temperature be in kelvins?
(A) 152.3 C (B) -120.7 C
(C) 152.3 K
(D) 162.5 K
Students in a Lab ran several tests and determined the statistics in the table below
for 4 unknown compounds. Use this table to answer questions 7-9.
Compound
Melting Point
Solubility in H₂O
P
Vap
W
176K
Low
40mm Hg
X
954 K
High
.32mm Hg
Y
86 K
Low-Medium
6400mm Hg
Z
1048 K
High
.01mm Hg
7. In Compound X which types of IMFs are most likely present?
I.
Dipole-Dipole
II.
London Dispersion Forces
III.
Ion-Dipole
A.
B.
C.
D.
I
I and II
III
II and III
8. Which compound(s) would dissolve in Octane?
I.
W
II.
X
III.
Y
IV.
Z
A.
I and III, because octane is polar.
B.
I, because it is the least soluble in H₂O and therefore nonpolar.
C.
I and III, because octane is non-polar.
D.
None of the compounds would dissolve.
9. Knowing KI is one of the unknown compounds, which compound would most likely
represent CaCl₂?
A.
W
B.
X
C.
Y
D.
Z
Use the table to answer questions 10-12. (it was assumed that these gases were in
a mixture)
gas
mass percentage
CO2
32.70%
N2
27.50%
He
9.40%
O2
30.40%
10. In a 1.5L container containing the above gasses at 32°Cat 928 mmHg, which gas is
most likely to break the container ?
A. Carbon dioxide
B. Nitrogen gas
C. Helium
D. Oxygen gas
11. The mixture is cooled until a liquid forms in the container: Which gas is it?
A. Carbon dioxide
C. Helium
B. Nitrogen gas
D. Oxygen gas
12. The mixture is heated back to a gas and transferred from the container into a
manometer. However, in the process, 36% of the gas escaped. Assuming that equal parts
of the moles of the gasses escaped and the outside environment is at standard pressure,
what would the manometer read?
A. 594 mmHg
C. 788 mmHg
B. 334 mmHg
D. 612 mmHg
Use the following table to answer questions 13 and 14.
A mixture at a constant volume, temperature, and pressure is composed of
chlorine gas, carbon dioxide gas, and oxygen. A table of the properties of the
mixture is provided below.
Moles of components
Partial Pressure (atm)
Cl2
?
2.21
O2
0.52 moles
?
CO2
0.45 moles
?
Mixture Total
1.4
?
13. Based on the table above, what would be the approximate total pressure of
the mixture?
A. 1.4 atm
C. 6.6 atm
B. 7.2 atm
D. 8.0 atm
14. When the mixture is in a container that is 3000.0 mL in volume, what is the
temperature in Kelvin?
A. 188 K
C. 188000 K
B. 273 K
D. 57.7 K
15. Below are experimental results from a student lab.
Based on only the factors provided, which of the following
would likely have the highest solubility in water?
State
Temperature
Pressure
Volume
A
Solid
Low
High
Medium
B
Gas
Low
High
Medium
C
Solid
Medium
Low
Medium
D
Solid
Medium
Low
Medium
16. Indicate which step(s) is or are occurring at parts A, B, and C
on the diagram
a. Solid à gas
b. Solid à liquid
c. Liquid à gas
d. Liquid à solid
e. Gas à solid
f. Gas à liquid
17.Which
of the following is the best representation of a calcium
nitrate solution in water?
PbCO₃→PbO+CO₂
PbCO₃ decomposed into PbO and CO₂, below the figures show the conditions of the
systems. Note, the temperature is held constant until PbCO₃ reaches the equilibrium with
PbO+CO₂.
PbCO₃
Ptotal= 1.00 atm
figure1: Initial
PbCO₃, PbO, and CO₂
Ptotal= 1.50 atm
Figure 2: Equilibrium
18. Which of the follow is the best explanation to why the pressure increases as the reaction
goes through decomposing and reaches equilibrium.
A.
The reaction produces heat which causes the molecules to move faster.
B.
An increase of molecules, which would increase the frequency of molecule collisions
against the wall of the container
C.
A decrease in the strength of intermolecular between separate molecules in the
container
D.
An Increase in the speed of the molecules, which causes the molecules to hit the walls
of the container with a greater force.
Use information below to answer questions 19 and 20
The vapor pressure of propane at 21.1 °C is 853.16 Kpa
Question 19
How many moles of propane will be in 1.00 L flask? (correct choice not present)
A.
.0038
B.
1.03
C.
.30
D.
.72
Question 20
The flask containing the propane is opened and more propane is added to the flask, the new
pressure rose to 884.3Kpa and the temperature is now 26.0°C, how many liters is the new
flask?
A.
.1L
B.
.82L
C.
1.4L
D.
.03L
21. Substance A dissolves in Silver Nitrate.
Substance A is found in a mixture and all its characteristics are unique to the other
substances in the mixture. When the mixture is passed through a chromatography column
with non polar lining when does substance A exit the column relative to the other
substances in the mixture? (This question may be referring to the table in #23 as the
components of the mixture – not sure)
A. First
B. Second
C. Third
D. Never
22. Which type of inter molecular force does substance A from the previous question
NOT have?
A. Van Der Waals force
C. London dispersion force
B. Dipole Dipole force
D. none of the above
23.
Melting Point
Substance A
?
CCl4
-22.92 C
C4H10
-140 C
C6 H6
5.5 C
Based on the table above and your knowledge of Substance A from the previous
questions, what is likely the melting point of Substance A?
A. 5.0 C
B.-150 C
C. 32 C
D. 5.5 C
A student conducted a gravimetric analysis of a compound containing only an
alkali metal element and the element iodine. The student reacted the unknown
compound with excess lead (ii) nitrate and collected the following…
Mass of unknown used: 2.964 g
Mass of precipitate produced: 5.103 g
Equation of Reaction: 2 XI (aq) + Pb(NO3)2 (aq) -> 2 XNO3 (aq) + PbI2 (s)
24. Would you predict that the unknown compound have a higher or lower boiling
point than potassium iodide?
a. Higher; the unknown has a higher molar mass than potassium iodide
b. Higher; the unknown has more intermolecular forces than potassium iodide
c. Lower; the unknown has less intermolecular forces than potassium iodide
d. Lower; the unknown has a lower molar mass than potassium iodide
25. The student attempts to explain what happens at a molecular level when the
unknown compound reaches its boiling point and becomes a gas. The best
explanation is…
a. Bonds within the molecule are broken
b. The molecule loses mass and therefore becomes lighter and forms a gas
c. Bonds between molecules are broken
d. Pressure forces an increase in volume and the molecules move apart to
become a gas
26. The student attempts to predict the temperature of the unknown compound in
a gaseous state when there are 46.84 g of the compound that occupies a volume
of 19.36 L at a pressure of 86.4 kPa. What should be the temperature of the gas?
a. 285 C
b. 300 C
c. 315 C
d. 330 C
27. Hot air balloons use large amounts of propane to fuel the hot balloons, which make
them highly flammable.
Write the balanced chemical equation for a hot balloons.
(a) C3H8 + 5O2 à 3CO2 + 4H2O
(b) C3H6 + 4O2 à 3CO2 + 2H2O
(c) C3H8 +2H2O à CO2 + O2 +5H2
(d) C3H6 + 4O2 à 3CO2 + 2H2O
28.Which of the following correctly explains how hot air balloons rises?
(a) As volume of the hot air balloon increases, the temperature increases. As the
temperature increases, density decreases. Hence, the hot and less dense air inside the hot
air balloon keeps the balloon afloat in the colder, denser air.
(b) As volume of the hot air balloon increases, the temperature decreases. As the
temperature decreases, density decreases. Hence, the hot and less dense air inside the hot
air balloon keeps the balloon afloat in the colder.
(c) As volume of the hot air balloon increases, the temperature increases. As the
temperature increases, density increases. Hence, the cooler air inside the hot air balloon
keeps the balloon afloat in the warmer air outside.
(d) No Gas Laws apply- only the altitude of the balloon makes a difference.
29. Which
of the following shows the association between pressure
(x- axis) and volume (y- axis) for situation above (assuming STP)?
The following questions will be referred to as #’s 30-32 (they
could not be edited)
33. Jim
combusts 60.0g of 3-Heptanol with 80.0g of Oxygen and wants
to collect Carbon Dioxide over Water at 10 degrees celsius. The
pressure in the system is 720 mmHg.
What volume of CO2 is collected?
A. 41.4L
B. 40.9L
C. 89.9L
D. 88.7L
34. Using the same temperature as above how fast on average does the
CO2 molecules travel compared to the 3-Heptanol molecules?
a. 2.5x faster
b. .39x faster
c. same
d. .15x faster
35. Jim later wants to test out Column Chromatography and puts a
solution of 3-Heptanol through a column that is coated in the middle
polar and nonpolar substance. What time should this solution go
through compared with a polar substance and a nonpolar substance?
(not exactly sure what this question is asking)
a.Slower than the polar but faster than the nonpolar
b.slower than the nonpolar but faster than the polar
c. Faster than the two
d. Slower than the two
Questions 36 through 38 refer to the following situation
Two different liquids are in a 5 L container. They are separated as shown. After
heating the container, an amateur chemist found that liquid B has a slightly
higher boiling point. Chemicals A and B are the only substances in the container.
36) Which diagram most likely reflects the particle diagrams of A and B before
boiling?
37) Ignoring space occupied by the liquids, how does the partial pressure of the
gaseous A change from the point when all of liquid A has vaporized and none of
B has vaporized to the point where all of both gases has vaporized?
A) Decreases
B) Remains Constant
C) Increases
D) Not Enough Info to Determine
38) Based on the Ideal Gas Law, which chemical, as a gas, would be more likely
to exhibit a molar volume closer to 22.4 L/mol, and why?
A) A, because A boils before B. This leads to a period of time when it is the only
gas in the container, leading to a greater average volume.
B) A, due to a lower boiling point, the molecules are likely nonpolar, thus less
attracted to each other and more ideal.
C) B, due to a higher boiling point, the molecules are likely nonpolar, thus less
attracted to each other and more ideal.
D) B, due to a higher boiling point, the molecules are likely polar, thus there is
more attraction between the gaseous particles and the mixture is more ideal.
39. An unknown gas was collected over water at 25 degrees celsius and 2.04
atm, and has a volume of 2000.0 mL and a mass of 11.6 g. Water vapor
pressure at 25 degrees celsius is 23.8 torr. Afterwards, the dry unknown gas was
added to a container holding 5.92 grams of neon gas, 3.18 grams of helium gas,
and 6.03 grams of oxygen gas with a total pressure 865 mm Hg (without the dry
unknown gas). What is the mole fraction of the dry unknown gas?
a. 0.117
b. 0.114
c. 0.119
d. 0.110
40 What is the identity of the unknown gas from question 1?
a. argon gas
c. chlorine gas
b. bromine gas
d. krypton gas
41. Which of the following alcohols will be most soluble with water: Ethanol, Butanol,
Propanol
a.
All the same; all three contain one –OH group and therefore have equal polarities
b. Propanol; the least polar molecule as it contains the most Carbons, will dissolve best
with water, which is polar
c.
Butanol; the most polar molecule as it contains the most Carbons, will dissolve best
with water, which is polar
d. Ethanol; contains the fewest Carbons, and is therefore the most polar, and will
dissolve best in water, which is polar
42. What accounts for the large jump in boiling point between H2S and H2O?
a.
H2O has the smallest molar mass
b. H2O contains intermolecular “hydrogen bonds”, which are stronger than other
dipole-dipole forces in H2S and H2Se
c.
H2O contains chemical bonds between hydrogens, or “hydrogen bonds,” which
must be broken to boil
d. H2O contains greater London Dispersion Forces than H2S and H2Se, and is therefore
harder to boil and separate the molecules
Compound
Boiling Point (standard atmosphere)
H2 O
100 °C
H2 S
-60 °C
H2Se
-42.3 °C
43. Which of the following particulate diagrams correctly displays a solution/mixture of KCl, Butane, and water a. b.
c.
d.
44.
Test Tube A:
Test Tube B:
1 mL of H O
1 mL of H SO
Final Temperature: 25 degrees Celsius
Final Temperature: 25 degrees Celsius
2
2
4
Based on the data above, what will the number of moles in Test Tube A be,
compared to Test Tube B at final temperature?
A.
Test Tube A will have more moles because there is a greater difference from
Final temperature and Initial Temperature.
B.
Test Tube A will have less moles because there is more grams per mole in an
H SO molecule.
2
4
C.
Test Tube A will have the same number of molecules as Test Tube B
because Temperature and Volume are constant, Moles stay constant.
D.
Answer cannot be determined from information given.
45
Test Tube A:
Test Tube B:
25 grams of Ch OH
3
25 grams of Ch Ch Ch Ch Ch OH
3
2
2
2
2
Based on the data given which test tube will be more soluble in 10 mL of H 0?
2
A.
Test Tube A will be more soluble because the polar and nonpolar parts of the
molecule is more balanced then Test Tube B.
B.
Test Tube A will be more soluble because the molecule has an overall
weaker bond than Test Tube B.
C.
Both Test Tubes are equivalent in dissolvability because there is the same
number of grams for both.
D.
Answer cannot be determined from information given.
46
Based on the diagram above, what compound does this image represent and
why?
A.
NaCl because the attraction is between a Cation and Anion and the force
being done here is ionic.
B.
NaCl because the electrons are being delocalized and the attraction has
cations.
C.
Fe because the electron sizes are the same for both positive and negative.
D.
Answer cannot be determined by information given.
2
The following is used to answer questions 47-49.
A nonpolar column is used for column chromatography when separating a solution of four
compounds, A, B, C, and D all dissolved in water. As a result of using the column, it is found that
the eluting time for A was the shortest, followed by B, and C, and finally compound D had the
longest eluting time. Also, it is found that A appeared in the solution at the highest molar
concentration, and B with the lowest. It is important to note that this test was conducted by a
student, and there could easily be errors in the data.
47) Which of the compounds would most likely have the highest boiling point?
a. A
b.
B
c.
C
d.
D
48) From the data, how can you see that A most likely has the highest solubility in water?
A. A has the highest concentration which shows it can easily dissolve in
water
B. A went through the tube the fastest so it must have the least amount of
nonpolar bonds
C. A went through the fastest so it must have hydrogen bonds like water
D. A has the highest concentration which is reflective of its weak
intermolecular forces.
49) Both compounds B, and D are found to have the chemical formula C3H8O - how do they
most likely have different eluting times?
A. The student tilted the chromatography column at random times, and angles during
the experiment.
B. One of the compounds is an ether, while the other is an alcohol
C. One of the compounds is a ketone, and the other is a ether
D. The student accidentally used a highly polar column instead of a non polar column.
50. Use the following figure to answer questions 50-52:
The compound in Figure 1 is liquid water and the compound in Figure 2 is hydrogen sulfide gas.
50. Based on the compounds given and the diagrams above which of the two compounds has
stronger intermolecular forces?
A. The water, because it is a liquid while the hydrogen sulfide is a gas.
B. The water, because it exhibits hydrogen bonding.
C. The hydrogen sulfide, because it exhibits hydrogen bonding.
D. The hydrogen sulfide, because it exhibits dipole-dipole forces.
51. Which of the following intermolecular forces are present in Figure 1?
I. Dipole-Dipole
II. Hydrogen Bonding
III.
Van der Waals Forces
IV.
London Dispersion Forces
A. I, II, IV
B. I, III, IV
C. II, III, IV
D. I, II, III, IV
52. Which of the following intermolecular forces are present in Figure 2?
I. Dipole-Dipole
II. Hydrogen Bonding
III.
Van der Waals Forces
IV.
London Dispersion Forces
A. I, II, IV
B. I, III, IV
C. II, III, IV
D.
I, II, III, IV
53. Which diagram best represents the situation the solution when zinc (Ⅱ) bromide and
potassium carbonate reacts completely (no
A.
B.
reactants left)?
C
D.
54. Of the following compounds, which option has higher boiling point than the latter?
A. MgCl2 and CaCl2
B. I2 and Br2
C. O2 and Ag
D. NH3 and CaO
55. Using the chromatography graph above, which of the following choices is correct? (the type
of column was not specified)
A. Substance A is more soluble in butane than substance B
B. Substance B is more soluble in water than substance C
C. Substance B has a higher concentration than substance C
D. Substance A is as soluble in water as substance C
56. The ratio of the number of grams of Carbon Dioxide to the number of grams of
Oxygen gas in a container is 7:3. If the total pressure of the mixture of the gasses is 4.20
atm., what is the partial pressure of the Oxygen gas?
A.)
B.)
C.)
D.)
1.26 atm.
1.56 atm.
1.80 atm.
2.28 atm.
57. In Mr. HD's alternate universe, pressure varies directly with both volume and
temperature. If, in this universe, 120. g of carbon dioxide at a pressure of 1.30 atm and
a temperature of 34.0 degrees Celsius occupies a volume of 1.20L, what volume would
7.00 g of carbon dioxide occupy at a pressure of 2.90 atm and a temperature of 52.0
degrees Celsius?
A.) .000117
B.) .102
C.) .148
D.) 1.22
58. While (which?) of the following solutions could be represented by the particulate
diagram below:
A.) water and potassium fluoride
B.) sulfur dioxide and potassium fluoride
C.) water and sodium chloride
D.) sulfur dioxide and sodium chloride
ANSWERS
1. Choice A is incorrect, as the molecular size does not address the polarity of the
molecules.
Choice C is incorrect, as it fails to address polarity and brings up an irrelevant aspect of
Butanol.
Choice D is incorrect, as the molecules differ in polar percentage of surface area. This is
a distractor meant to catch students who do not fully understand polar to total surface
area ratios.
Choice B is correct because it accurately describes the higher ratio of polar surface area
to each molecule’s total surface area. For this reason, it will dissolve more easily in the
polar solvent, water. Correct
Choice A is incorrect because it assumes heat of vaporization for water changes with
mass.
2. Choice C is incorrect because greater surface area would cause less energy to come to
a given molecule per unit of time, as the energy would have to be spread evenly
Choice D is incorrect, and also a distractor. It is meant to be appealing to students who
have memorized that the equal vapor pressures correspond to equal boiling points.
However, it ignores the dependence of time to boil on the mass of water present, as it
would require more energy to heat up a greater mass of water to boiling.
Choice B is the correct answer, as it accurately addresses the dependence of time to boil
on the mass of water present. Container A would require less heat energy to completely
boil all of its water than container B. not sure about this – assumes the water in both
containers is already at 100 deg C to have that high of a vapor pressure
3. Choice A is incorrect because it is the equivalent measure of Choice B in atm, and
cannot be correct without B being correct as well.
Choice B is incorrect because it is Choice A’s equivalent in torr rather than atm, and it
cannot be correct without A being correct as well.
Choice D is incorrect, and is this question’s distractor. Students attempting to shortcut the
problem by seeing the temperature doubles in centigrade and applying the same change
to pressure ignore the need to use Kelvin temperatures for gas laws. As a result, the
answer for choice D is grossly exaggerated.
Choice C is correct, as it validates both A and B, the two equivalent and correct values
for the new pressure. Correct – good distractor that many students might choose
4. B. Ammonia : the only compound with a hydrogen bond
Most people wouldn't know the formula for formaldehyde so that would have
them hung up on it. CORRECT – note that NaCl does not have intermolecular
forces as it is not a molecule!
5 - D. 2660 mm Hg : 1.6mol/4.0atm = 1.4mol/x
(1.6mol)x = (1.4mol)(4atm)
x = 3.5 atm
•
(3.5 atm)(760 mm Hg/1 atm) = 2660 mm Hg
• the 3.5 torr is there to distract the people who wouldn’t read the full
answer - torr instead of atm even though the number is the same
CORRECT
6 - C 152.3 K : PV=NrT
(10.5 atm)(5 liter)=(4.2 mol)(.0821)(t)
t = (10.5)(5 liter)
(4.2 mol)(.0821)
•
t = 152.3 K
A is to distract non-readers of the question, and B is to distract those who got the
answer in celsius since the answer should be in Kelvin CORRECT
7. B, this is because with compound X’s high melting point and solubility in water it is
an Ionic solid most likely a salt. Salts themselves have dipole-dipole IMFs and all
compounds have LDFs. Agree that it is probably ionic - however ionic compounds do
not have any IMF (there are no molecules)
8.
C, this is because compound W and Y with their low melting points
and low-medium solubility in water must be nonpolar
molecules. Octane which is also a nonpolar molecule will dissolve like
molecules. Compound Y represents an organic molecule with a small
amount of carbons making it slightly soluble in water while compound
W is a high carbon compound being less soluble in water. CORRECT
- note that Y cannot have that high of a vapor pressure! (760 is the
max?)
9. D, this is because while both X and Z compounds are ionic compounds, and Z has
a higher melting point. You know KI and CaCl₂ are unknowns and because of Lattice
NRG CaCl₂ will have a higher melting point. CORRECT
C: Heliumbecause it has the greatest partial pressure of the gasses.
CORRECT - (since V,R and T are constant the greatest # of moles will produce
the largest P – if we assume there are 100 g of mixure – He has 2.35 moles and
the rest are under 1 mol)
10.
11. A: Carbon dioxide has the strongest LDFs of the gasses, making it take less energy to
condense it. CORRECT
12. A: Equal parts of the gasses escaped, meaning a decrease in pressure, making as
though all the gasses mixed in are one gas with a 36% decrease in moles. Thus, only the
pressure from question 1 needs to be dropped 36% to get the answer. Unlikely equal
moles of each gas would escape as they each diffuse at different rates – more CO2 would
be left than any other. However if equal moles did escape, this could be correct.
13. The correct answer for question one is B. 7.2 atm. The total mixture pressure
could easily be calculated by Dalton’s Law of Partial Pressures.
Partial pressure = mole fraction X total pressure, so dividing partial pressure by a
component’s mole fraction would yield the total pressure.
The chart provides the partial pressure. The moles of Cl2 can be found by
subtracting the moles of CO2 and O2 from the total number of moles. Therefore:
Total pressure = 2.21/ (.43/1.4) = 7.2 atm. Choice A is incorrect because it is just
the sum of moles of the mixture in atm. Choice C is incorrect because all it is
doing is taking the provided partial pressure of Cl2 and multiplying by 3. Choice
D is incorrect because it is merely taking the total moles and adding it to choice
C. CORRECT – nice question
14. Using PV=nrT, the student can set T so that T=PV/nr. The total pressure was
calculated in part 1. The volume given in mL can be converted to 3L. The total
moles and gas constant are also available. Therefore:
T=PV/nr= (7.2 x 3)/(1.4 x .08206)=188K
The correct answer is choice A. Choice B is wrong because it is merely the
temperature of a gas at STP. Choice C is wrong because the volume wasn’t
converted to Liters before using the ideal gas law. Choice D is wrong because it
is using the partial pressure of Cl2 instead of the total pressure of the mixture
and substituting it into the ideal gas law. CORRECT
15. B would have the highest solubility in water. As a gas, it has a low temperature, which
indicates that there is less energy in the molecules. A higher kinetic energy makes it too easy for
the gas to leave its phase and become liquid. The gas also has a high pressure, and partial
pressures above the pressure of the solution make the gas more soluble, according to Henry's
Law. Henry's Law states that the solubility of a substance is directly proportional to the partial
pressure of a gas. The volume remains constant in this case and is not a
consideration. PARTIALLY CORRECT – the idea of a gas being more soluble in water at low T
is right, however it is difficult to tell with the solids – what is they are ionic? (and hence v soluble
in water). Also the application of Henry’s Law is only is a closed container for the vapor
pressure above the solution.
16.
A: This is the point of both sublimation (a, solid to gas) and deposition (e, gas to solid)
because both the pressure and temperature are low.
B: This is the point of both vaporization (c, liquid to gas) and condensation (f, gas to liquid)
because the temperature is high but the pressure is still relatively low.
C: This is the point of melting (b, solid to liquid) and freezing (d, liquid to solid) because both
the temperature and pressure are high. Agree with these – however this is not really a MC format
question – not sure what they water dashed line means on the diagram as well.
17. A is the best representation of a calcium nitrate solution in water. The calcium cation is
smaller than the nitrate anion in terms of atomic radii since the nitrate ion is composed of one
nitrogen and three oxygen atoms. On the other hand, the calcium is on the left upper side of the
periodic table which means that its valence electrons occupy a low and smaller energy level. On
the calcium, the water molecule should have its hydrogen closer to the calcium and its oxygens
facing away from the calcium. The hydrogen should be farther away from the nitrate ion and the
oxygens should be closer to the nitrate ion. This is because hydrogen has an overall positive
charge, and is attracted to the positive cation. The oxygens have an overall negative charge, and
are attracted to the negatively charged anion. CORRECT
18. B Explanation: Since the compound is going through a reaction and decomposing it causes
the container to have more molecules, and by having more molecules the amount of times that the
molecules are colliding with the sides of the contain increase; which causes the pressures to
increase. CORRECT - however this only applies to gas molecules (carbon dioxide in this case)
19 C Explanation: By setting up the equation PV=nRT you get the outcome of .3 for n; to do this
you also have to convert specific units into others for example Kpa- atm (conversion not needed
if using R = 8.31), Answer = 0.35 mol
20.A Explanation: By using the .3 for n that was calculated in the previous question its a simple
gas law equation to figure out the new amount of liters. (the moles have increased - so this
problem cannot be solved as given)
21. A
silver nitrate is polar therefore substance A is polar because like dissolves like. all characteristics
of substance A are unique to the other substances in the mixture so substance A is the only polar
substance. It will pass through the column fast since the column is non polar- lined, and therefore
will pass through first in relation to the other substances. (CORRECT – would help to use table
from #23)
22. D
as we know substance A is polar. van der waals force is all inter molecular forces which
substance A contains. Dipole dipole forces result from polar substances, therefore substance A
has them, and all substances have London dispersion forces, so none of the forces listed are not
present in substance A. CORRECT – many people confuse Van Der Waals forces as being the
same as LDF – they are correctly defined here
23. C
the more polar a substance the more attraction between molecules and therefore the higher
temperature must be for the substance to turn from a solid to a liquid, or melt. the other
substances in the table are all non polar, and we know from previous questions substance a is
polar, so it should have a higher melting point that the rest of the substances. Not necessarily –
LDF can be quite strong as an aggregate – this question will depend on the mass of substance A.
If A is similar is mass to the heavier nonpolar substances, it will be higher. If A is significantly
lighter – than it will be lower than the nonpolar substances.
24. The unknown was calculated by first finding the moles of precipitate produced. Then using
stoichiometry and the chemical equation for the reaction, the number of moles of unknown used
was calculated. Then a proportion was set up for the number of calculated moles used and the
mass of the unknown used and 1 mol of the unknown and the the molar mass (calculated). The
mass of chloride was then subtracted from the molar mass and the using the answer and
comparing to elements on the periodic table, the unknown was identified.
b- The unknown compound came out to be Lithium Iodide and because there is a higher
difference in electronegativities in LiI there are more intermolecular forces that must be broken
making the bp higher. The ID is correct – however ionic compounds do not have intermolecular
forces as there are no molecules – a better reason is Coulumb’s law
25. c- Only intermolecular forces are overcome during phase changes. The question focuses on
what happens during a phase change. It is important to know that only intermolecular forces are
overcome and not intramolecular forces (answer letter a) and answers b and d are incorrect
because mass and pressure are not relevant in boiling point. CORRECT – if there are
intermolecular forces – LiI is ionic and has a crystal lattice so lattice NRG comes into play
26. b- Using PV = nRT the temperature came out to 575 K which is equal to 300 C
P= 86.4 V= 19.36 n= 0.35 (based on molar mass of LiI) R= 8.31 T= ? CORRECT
27. Correct answer: C3H8 + 5O2 à 3CO2 +4H2O
Knowing the formula for propane, students should be able to get C3H8, not C3H6 which
eliminates choice (b) and (d). From prior knowledge, C3H8 is a hydrocarbon which means that it
will create carbon dioxide and water as products. The description above mentions that the hot air
balloons are highly flammable so the equation is a combustion reaction. Hence, oxygen will also
be a reactant with propane, which will be used to produce carbon dioxide and water (C3H8 +
O2 à CO2 +H2O) which eliminates (c). Last step is to balance the equation which would make
the final balance equation C3H8 + 5O2 à 3CO2 +4H2O, option (a) CORRECT
28. (a) According to the Charles’ Law, (V1/T1=V2/T2), as the temperature of the hot air balloon
increases, the volume also increases. For the next step, students have to use the alternate form of
the Ideal Gas Equation, PM= dRT. However, as the temperature increases, the density decreases.
For this reason, the air inside the hot air balloon is hot and less dense than the cold, and denser air
outside. CORRECT
29. Correct answer: (a)
Since the gas is at STP, this question requires the Ideal Gas Law (PV= nRT). Temperature,
volume, moles the Gas Constant (0.0831 L bars/Kmoles) can be disregarded as they are just
constants. Since volume and pressure are inversely related, (b) and (d) cannot be the answers as
they show direct relation. Choices (a) and (d) both show inverse association but choice (a) shows
negative inverse association while choice (d) shows positive inverse association. The correct
answer is (a) because as pressure increases, volume decreases. CORRECT
30.
A. RMS is inversely proportional to molar mass. In order to solve this problem, the molar mass of
each compound should first be determined. Since the temperature, pressure, and volume are
given, the ideal gas law must be used.
Compound W: Partial Pressure = 1 atm - partial pressure of oxygen PP = 1 - 0.6 = 0.4 atm
n = (PV)/(RT) molar mass = (mass * R * T)/(PV) M = (0.419 g * 0.08206 * 373)/(0.4 atm * 1 L)
= 32 g
Compound X: Partial Pressure = 1 atm - partial pressure of oxygen PP = 1 - 0.78 = 0.22 atm
n = (PV)/(RT) molar mass = (mass * R * T)/(PV) M = (0.331 g * 0.08206 * 373)/(0.22 atm * 1 L)
= 46 g
Compound Y: Partial Pressure = 1 atm - partial pressure of oxygen PP = 1 - 0.846 = 0.154 atm
n = (PV)/(RT) molar mass = (mass * R * T)/(PV) M = (0.362 g * 0.08206 * 373)/(0.15 atm * 1 L)
= 72 g
Compound Z: Partial Pressure = 1 atm - partial pressure of oxygen PP = 1 - 0.8 = 0.2 atm
n = (PV)/(RT) molar mass = (mass * R * T)/(PV) M = (0.380 g * 0.08206 * 373)/(0.2 atm * 1 L)
= 58 g
Since compound W has the smallest molar mass, it has the fastest rms. Choice Y would be
tempting to students who believe that rms is directly proportional to molar mass. Note: Rms
velocity does not need to be calculated for this problem. CORRECT – interesting question
31. C. First, the formula for each compound needs to be determined.
Compound W Since the partial pressures of water to carbon dioxide are at a ratio of 2:1, the ratio
of hydrogen to carbon in the compound is 4:1 because there are two hydrogen atoms in water and
one carbon in carbon dioxide. Since the molar mass of the compound is 32, the number of carbon
and hydrogens can be solved for using the molar masses of carbon and hydrogen. The compound
would be CH4O.
Compound X Using a similar method as compound W, the molecular formula of X is C2H6O
Compound Y The molecular formula is C4H8O
Compound Z The molecular formula is C3H6O
Out of all substituted hydrocarbons with one oxygen, alcohol compounds have the highest boiling
point because it is the only one with an oxygen (an extremely electronegative element) bonded to
a hydrogen. Due to this, it is able to have hydrogen bonds with itself. Hydrogen bonds are the
strongest type of dipole-dipole interaction, so the intermolecular forces between the molecules are
strengthened. Compound W: methanol CH3OH
Compound X: ethanol CH3CH2OH
Compound Y: 2-butenol CH3CH2CHCHOH
Compound Z: 2-propenol CH2CHCH2OH
Since all 4 compounds can be written as alcohols, the Van Der Waals forces should be
considered. The larger the molecular weight, the more polarization, and therefore the
intermolecular forces would strengthen resulting in a higher boiling point. Since compound Z has
the highest molar mass, it would therefore have the highest boiling point. A student that would be
misinformed about the relationship between mass and boiling point would pick A because it has
the smallest molar mass, which is incorrect. CORRECT – very interesting question and hard one
to figure out; good explanation
32. A. Compound X is ethanol, so it has two carbons and six hydrogens. Diagram A and C are the
only ones that have ethanol, and only in A is the water oriented correctly based on polarity. The
diagram in C would be tempting to students who do not understand that water is more negatively
charged at its oxygen and positively charged at its hydrogens. Choice B is a correct representation
of methanol, which is compound W. Choice D is a correct representation of propenol, which is
compound Z. CORRECT – good diagram and explanation.
33. The answer is A. First the student must write a balanced chemical equation then proceed to
use the mass of heptanol provided to obtain moles of O2 needed. Then, test out with the mass of
Oxygen to find out how much the equation has. Once this is obtained the student must uses the
Limiting reagent O2 and find out moles of CO2. After this is done, the student can proceed to the
Pv=NRT and solve for volume. They must use the right constant and also the right pressure and
moles. Student cannot use in celsius and must convert to kelvin. NOTE: The student must find
out the pressure of CO2 in order to complete the final step. To do this the student needs to know
the constant of water which is 9.2 mmHg and subtract that from total and find Pressure of the
CO2. CORRECT – note that if one uses Heptanol as the LR, the answer of about 90 L (choices C
and D) appear)
34. The answer is B. In order to answer this question correctly, the student needs the Grahams
Law of Diffusion. Once the student knows that they need this law, they also must remember to
obtain molar mass. Also the student must remember how to draw 3-Heptanol and figure out the
formula. A common mistake is not realizing the ratio is CO2 to 3-Heptanol therefore they might
pick answer a. quickly. Actually the answer should be 1.62 (square root of 116/44).
35. The answer is C. In order to answer this question correctly, the student must remember what
column chromatography does and how it functions. The student must know if the column is
coated with nonpolar or polar substances, what effect will it ensue in with the time it takes to
come out. In this case 3-Heptanol has both polar and nonpolar components to it. The student
might think that because it has both parts one might happen faster than the other. This is not the
case because it must be overall which one comes out faster. Therefore, 3-Heptanol should take
the fastest time to run through. Believed to be correct, based on the wording that the column had
“middle polarity”, best matching that of 3-heptanol.
36. A) is correct. B has a higher boiling point. This indicates a likely polar nature between
the molecules. This is seen with the molecules facing the same direction. A is likely
nonpolar due to the lower boiling point. This is apparent with the randomly aligned
molecules. B) is a distracter because the molecules appear to be water, which is polar.
However, as shown, these molecules are not water due to the nonpolar nature.
CORRECT
37. B) is correct. The pressure of the gaseous B remains constant even when additional
gas is added. Only the total pressure changes. C) is a distracter, because one would
expect the pressure of the gas to increase as more gas is added to the container.
CORRECT
38. B) is correct. The likely nonpolar nature of A means there is less attraction between
the particles. 22.4 L/mol is the the molar volume of an ideal gas, so a gas that is more
ideal is more likely to exhibit this molar volume. Less attraction between the particles
results in the gas being more ideal. C) is a distracter, because the justification is correct,
but the chosen gas is incorrect. CORRECT
39. B, there was a lot of extra information given in this problem, which would not be
needed until question 2. By using ideal gas law, one would be able to find the amount of
moles of the gas. The moles of the other gases in the container could easily be found,
and then would simply have the moles of the unknown gas divided by the total amount of
moles to get the mole fraction. It is possible that when using ideal gas law, students can
forget to subtract the water vapor pressure from the total pressure as the gas was
collected over water.
Math:
PV = nRT
P = 2.04 atm - (23.8 torr * 1 atm/760 torr) = 2.01 atm
(2.01)(2) = n(0.0821)(298)
n = 0.164 mol unknown
5.92 g Ne * (1 mol/20.1797 g) = 0.293 mol Ne
3.18 g He * (1 mol/4 g) = 0.795 mol He
6.03 g O2 * (1 mol/ 32 g) = 0.188 mol O2
total moles = 0.164 + 0.293 + 0.795 + 0.188 = 1.44 moles
mole fraction of unknown gas = 0.164/1.44 = 0.114 CORRECT
40. C, the second version of ideal gas law, PM = DRT, would most likely be used to
solve this problem. If students forget to convert 2000.0 mL to 2.0 liters or that it is
possible for some of the choices listed to be diatomic, answers may be off.
Math:
PM = DRT
P = 2.01
D = 11.6 g/ 2 L = 5.82 g/L
2.01(M) = 5.82(0.0821)(298)
M = 70.9
The molar mass most fits the gas which is Cl2,chlorine gas. CORRECT
41. d-­‐ ethanol has the fewest Carbons, and because polar compounds best dissolve or are most miscible with other polar compounds, and water is polar, ethanol is the correct response. a was meant to serve as a distractor, if the student thinks that only the number of polar groups (1 -­‐OH group) is what determines polarity. CORRECT 42. b-­‐ hydrogen bonds between water molecules is what causes the large jump in boiling point. c was meant to serve as a distractor; it uses the correct term but describes it as a chemical bond; IMF forces are not actually bonds. CORRECT 43. a-­‐ Butane does not mix with water, as it is nonpolar, and KCl dissolves in water, as it is polar. Also, K, which is larger than Cl, lines up with O from water, and Cl with H, as KCl is polar and will therefore dissolve in water. b and d were meant to serve as distractors. b has the whole solution evenly distributed, instead of the butane on top and water below. d has the incorrect number of Carbons, testing knowledge of organic chemistry. Interesting question – however Cl-­‐ is larger than K+ 44. The answer for this question is D. This question involves gas laws. From the
information given, you know that the compound, the temperature, and the volume of
each compound is given. This part should confuse the student a little, because they
cannot figure out how many moles there are without knowing the pressure. But in the
students mind they presume that the pressure stays constant. Choice C is the main
distractor because the choice states that since the temperature and the volume stays
constant than the number of moles should stay constant. Like said before, this would
confuse the student because they would presume that that the pressure is constant. A
few issues - however we also don’t know the concentration of H2SO4 - also neither of
these compounds are gases at 25 degrees C, so it would be hard to use gas laws to
answer the question
45. The answer for this question is A. This question has to do with polarity. If the student
understands chemistry, s/he would know that OH is the polar part of both molecules and
that the rest of both compounds is non polar. Since water is polar, this means that Test
Tube A would be more soluble because the polar and nonpolar side is more balanced.
Test Tube B is more non polar on the left side then on the right side. This means it is
slightly soluble. In this question the main distractor would be choice B. Here a student
who doesn’t know much about polarity might presume B because they look at the
structure and see that there are more bonds and that means B is stronger. Even they
are right about Test Tube A, the reason for it is incorrect, and would be marked wrong.
CORRECT - although it might be better to say “is more polar” than “balanced”.
46. The answer for this question is A. This question involves Cations and Anions. By
being able to read the diagram, the student should be able to realizes that the diagram
involves cations and anions, without needing to know that the force is ionic. The main
distractor is choice C. Here a person who is skimming through the choices, might think
that it has to be C because they might read it as the molecules, and think the atomic size
is the same for Fe2 but not for NaCl, therefore choosing option C. CORRECT - although
probably a bit too easy to solve.
47. The answer is A. From the data given, it is seen that compound A went through the nonpolar
column the fastest. This would mean that this compound would have adhered to the nonpolar
surfaces of the column the least and that therefore that the compound would have to be the most
polar in comparison to the other three compounds. As this compound is the most polar it will
have the strongest intermolecular forces. Therefore, it will take the most energy to break the
bonds between the molecules of this compound, and it will have the highest boiling point.
CORRECT
48) The answer is B. Water is highly polar due to its hydrogen bonds, therefore the most polar
compound will have the highest solubility. As compound A had the fastest eluting time, through
the non polar column this compound would be the most polar out of the four compounds. In the
possible answers, A and D are about concentration which would not influence solubility.
Although they can be used as a trick if the student did not understand the concepts as they are the
only answers directly dealing with solubility, and intermolecular forces. Answer C, is half right,
so in that sense it is a distraction, however there is no indication that there is a hydrogen bond.
Therefore, A is the correct answer. CORRECT
49) The answer is B. A, might influence the data however, it is not the best answer, and it is
highly unlikely that the compound would come out in two neat time frames if the column was
being used in the way described. D, would have made the data incorrect however, it would still be
testing for relative polarity and therefore if B and D were the same compound they would never
have eluted in separate time frames. The final answers, B, and C, are harder to distinguish
between as they both address the correct issue of the question, isomers. C is in answer choices to
make it harder to distinguish between the answers, however it is clear that C is wrong because the
proper chemical formula for the ketone would be C3H6O not C3H8O, the formula given. And
answer B does refer to two possible compounds,propanol, and methyl ethyl ether, and is correct.
CORRECT
50. The correct answer for this question is B. The water’s hydrogen bonding is the strongest force
named. Hydrogen bonding is stronger than regular dipole-dipole forces so D cannot be the correct
answer. Also, hydrogen sulfide cannot exhibit hydrogen bonding because hydrogen bonding only
occurs between hydrogen and either nitrogen, oxygen, or fluorine and therefore C cannot be the
correct answer. Lastly, the phase of a molecule involves intramolecular forces and not
intermolecular forces so A cannot be the correct answer. A student who did not know the
difference between intramolecular and intermolecular forces might have felt that that A was the
correct answer so this was the main distractor. Actually the phase of a substance depends on
intermolecular forces for covalent molecules (low IMF = gas) Otherwise correct.
51. The correct answer for this question is D. The water exhibits hydrogen bonding between its
oxygen and hydrogen atoms and hydrogen bonding is a form of dipole-dipole forces.
Additionally, London Dispersion Forces are always present due to being instantaneous dipoles.
Finally, Van der Waals forces encompases all intermolecular forces and since intermolecular
forces are present so are Van der Waals forces. The main distractor here was the listing of dipoledipole forces. If a student did not know that hydrogen bonds are a stronger form of dipole-dipole
forces they might have gone for C instead assuming that the hydrogen bonding occurred in place
of dipole-dipole forces. CORRECT
52. The correct answer for this question is B. The hydrogen sulfide exhibits dipole-dipole forces
between the hydrogen atoms and the sulfur atoms. Additionally, London Dispersion Forces are
always present due to being instantaneous dipoles. Finally, Van der Waals forces encompases all
intermolecular forces and since intermolecular forces are present so are Van der Waals forces.
The main distractor here was the inclusion of hydrogen bonding. If a student was not aware that
hydrogen bonding only occurs between hydrogen and either nitrogen, oxygen, or fluorine they
may have assumed sulfur could create hydrogen bonds. However, this is not possible and
therefore it is the only kind of intermolecular force not present in Figure 2. CORRECT
53. Correct answer： C
zinc (Ⅱ) bromide and potassium carbonate reaction:
ZnBr2(aq) + K2CO3(aq)→ZnCO3(s)+2KBr(aq)
When completely reacted, there would only be potassium bromide left in the liquid, since
ZnCO3 is insoluble in water and would precipitate. Potassium bromide is a salt and
therefore dissociates in water and is separated into potassium ions and bromine ions.
The relative sizes of the K+, Br—, H and O are: Br—> K+ > O > H. Potassium ions
(positive charge) are attracted the oxygen atoms of the water molecules and bromine
ions (negative charge) are attracted to the hydrogen. Therefore the only choice that
satisfies all above is C. In A, the potassium ions are attracted to the hydrogen atoms, the
bromine ions are attracted to the oxygen atoms. In B, the oxygen atoms are larger than
the bromine ions. D is just completely structurally incorrect, water molecules do not
dissociate. CORRECT
54. Correct answer : B
In B, the only intermolecular force involved for both of the nonpolar molecules is London
dispersion. And as I2 has more electrons and larger in size, it has a stronger
intermolecular force and thus a higher boiling point.
CaCl2 has a larger difference in electronegativity of bonded atoms than MgCl2, so A is
wrong. C is wrong, for solids always have a higher boiling point than gas. D is wrong
because CaO has ion-dipole forces while NH3 has diople-diole forces which is weaker
than ion-dipole forces. CaO has a larger difference in electronegativity than NH3, which
means a stronger intermolecular force, and therefore a higher boiling point. CORRECT
55. Correct answer: B
A chromatography column is made of non-polar materials, when a mixture goes through, the nonpolar substance takes longer eluting time and thus, comes out slower. Therefore the first to come
out is the most polar. In the problem, the ranking of polarity is: A>B>C. SInce polar substances
dissolve in polar substances, nonpolar substances dissolve in nonpolar substances, substance B
dissolves more in water than C for it is more polar and water is polar. Choice A should be
substance A is less soluble, and choice D should be substance A is more soluble than C in water.
For choice C, the concentration cannot be defined. CORRECT – if the column is nonpolar.
56. Even though only a ratio is given, the actual mass is irrelevant because mole fraction (in
other words a ratio) is used to find the Partial Pressure. We can therefore simply assume that
there are 3 grams of O_2 and 7 grams of CO_2.
For O_2: 3 g * ( 1 mol / 32.0 g) = .0938 mol O_2
For CO_2: 7 g * (1 mol / 44.0 g) = .159 mol CO_2
Partial Pressure of Oxygen gas = mole fraction * Total Pressure = ((.0938) / (0938 + .159)) * 4.20
= 1.56 atm. CORRECT
57. The new ideal gas law to be used should be VT = nRP (another can be used, but by the
context of the units of the answers, this is the best choice).
moles CO_2: 120. g x (1 mol. / 44.0 g) = 2.73 mol.
(1.20 L) x (273 + 34.0 K) = (2.73 mol.) x R x (1.30 atm)
R = 104 (L x K) / (mol x atm.) -------> new gas constant
(n L) x (273 + 52.0 K) = (7.00 g x 1mol / 44 g) x (104 (L x K) / (mol x atm.)) x (2.90 atm.)
n = .148 L Actually the new ideal gas law should be P = nRTV as T should not be
affected/moved. Interesting question however.
58. The solute was the easier of the two parts to choose. The two options were potassium
fluoride and sodium chloride. In the diagram, the cation is shown to be much smaller than the
anion. This is true for sodium chloride, but not for potassium fluoride because potassium is
larger than fluoride. This means that the correct solute would be sodium chloride.
The solvent in the solution can be either sulfur dioxide or water. Both have a bent shape and the
lone atom is larger than the double atom in both (ex. S is larger than O and O is larger than H).
This makes them difficult to differentiate. However, with sulfur dioxide, Oxygen has a higher
electronegativity meaning that the lone atom (sulfur) will have a positive charge as opposed to
water where oxygen has the higher electronegativity making it (the lone atom) have a negative
charge. Since the lone atom is attracted to the positive ion, it must have a negative charge making
the solvent water. CORRECT