Download WA AP Chem gas law IMF MC Set C

WA AP Chemistry
Exam 4 MC review Set C
Comments in red are teacher comments and not part of original work
1. 30.0g of CO2 in a rigid container is heated from 100 °C to 300 °C. Which of the
following best describes what will happen to the pressure of the gas? The
pressure will ...
A. decrease by a factor of three
B. increase by a factor of three
C. increase by a factor less than three
D. decrease by a factor greater than three
Four unknown substances were tested in the laboratory. A novice chemistry
student, whom has not performed many chemistry experiments, collected the
following data. There may be some errors to the data.
Substance
E
F
G
Molar Mass
95.211 g/mol
42.394 g/mol
58.44 g/mol
Boiling Point
low
low
high
H
110.98 g/mol
high
He knows that the substances are calcium chloride, sodium chloride, lithium chloride,
and magnesium chloride.
2. The student dissolves substance H in liquid H2O. What happens to the solution?
A. It becomes endothermic and feels cold
B. It becomes endothermic and feels hot
C. It becomes exothermic and feels cold
D. It becomes exothermic and feels hot
3. This situation refers to questions 3 and 4
An unknown gas at STP diffuses at the same rate as neon at -102℃ and 1.5
atm.
What is the identity of the gas?
A. Phosphorus B. Sulfur
C. Oxygen D. Nitrogen
4. How would the densities of the gases at these conditions compare?
A. They would be the same
B.
The density of neon would be more than twice of the density of the
unknown gas.
C.
The density of the unknown would be greater than that of neon.
D.
The density of neon would be greater than that of the unknown.
5. A student was given four solutions and asked to determine their identities
at the end of the experiment. The student ran many tests to help them do
this. These are their results:
Solution 1: Very soluble in acetone, soluble in water, boiling point of
120℃
Solution 2: Soluble in water, soluble in acetone, boiling point of -10℃
Solution 3: Insoluble in water, insoluble in acetone, boiling point of 130℃
Solution 4: Soluble in water, soluble in acetone, boiling point of 750℃
Which solution is most likely 1-butanol?
A. 1
B.
2
C.
3
D.
4
The following pertains to questions 6-8
An aqueous solution of 2g potassium carbonate reacts with excess hydrochloric
acid at 25°C and 770 torr (Vapor pressure of water is 23.8 mmHg at 25°C).
6. Which pure reactant has a higher melting point point?
A. Hydrogen chloride, because its London dispersion forces are greater
B. Hydrogen chloride, because its hydrogen bonding is greater
C. Potassium carbonate, because its crystal lattice must be broken
D. Potassium carbonate, because its London dispersion forces are greater
7) The gas produced in this reaction is collected over water. How many mL of
gas are produced?
A. 338mL
B. 349mL
C. 503mL
D. 488mL
8. How much concentrated (18M) hydrochloric acid is needed to complete this
reaction?
A. 0.81mL B. 1.12mL C. 1.67mL D. 2.24L
This refers to questions 9-11.
The Stöber process is an industrial manufacturing process in which tetraethyl
orthosilicate (SiC8H20O4) is reacted with excess water in the presence of several
catalysts to produce silica (SiO2) nanoparticles and ethanol, an alcohol that is miscible in
water. To perform the reaction, gaseous tetraethyl orthosilicate at 500 K and 3 atm is
piped into a tank containing excess water where it is condensed to initiate the reaction.
9. Following the reaction, a lab technician analyzes the nanoparticles and finds that they
have an average volume of 5.236*10^−1 µm^3. If 3.0*10^−12 moles of nanoparticles
were formed and all of the tetraethyl orthosilicate is consumed, what volume of gas was
piped into the reactor assuming silica has a density of 2.65 g/cm^3?
A. 4.1 * 10^−11 L
B. 4.2 * 10^−2 L
C. 5.7 * 10^−1 L
D. 3.4 * 10^1 L
10. Following this measurement, the nanoparticles are filtered out of the solution and the
catalysts are removed such that only water and ethanol remains. Given that ethanol has a
density of 0.789 g/cm^3, which of the following is true of the resulting solution?
I. Ethane would not dissolve readily in the solution
II. Whether ethanol is fully dissolved is dependent on the volume of water
initially in the tank
III. Ethanol’s molarity cannot exceed 17.13 M
A. I only
B. I and III
C. II and III
D. I, II, and III
11. In another run of the reaction and filtering process, one of the catalysts, propanol, is
not removed from the solution. Which of the following correctly orders the three
constituents of the final solution in order of increasing boiling point?
A. Propanol, ethanol, water
B. Water, ethanol, propanol
C. Ethanol, water, propanol
D. Ethanol, propanol, water
Use the following table to answer questions 12-­‐13 Unknown Compounds
Bond Type
W
Ionic
X
Non-polar Covalent
Y
Polar Covalent
Z
Non-polar Covalent
12. Rank the unknown compounds’ melting points in order from the lowest
temperature to the highest temperature.
a) X, Y, Z, W
b) X, Z, Y, W
c) Y, W, X, Z
d) W, Z, Y, X
Molar Mass
58.5 g/mol
28.0 g/mol
28.0 g/mol
70.9 g/mol
13. If all compounds were eluted through a stationary phase lined with non-polar
material, which compounds would you expect to come out of the column first?
a) X and Z
b) W and Z
c) W and Y
d) X and Y
14.
If 5.00 g of ethane was completely combusted in excess oxygen at an
atmospheric pressure of 3.00 atm and 65.0⁰C, calculate the volume of CO2 over
water. The vapor pressure of water is 187.5 mmHg.
a) 2.98 L
b) 4.60 L
c) 36.2 L
d) 3.25 L
The following answer choices are used to answer questions 15 and 16. Answers may be
used more than once
(A) H2O (B) SrCl2 (C) 2-heptanol (D) Benzene
__ 15. Which has the second highest boiling point?
__ 16. Which compound exhibits resonance in its lewis structure?
17. A first year chemistry student performed an experiment to determine the boiling
points of three different solids. The boiling point of the three solids he tested are listed
below. In what compound did the bonds break last upon boiling in the student's
experiment?
(A) An ionic compound - 740ºC
(B) A covalent network solid - 4,800ºC
(C) An amorphous solid - 899ºC
(D) None of these
Use the following data table for questions 18-20.
Substance
Methane
Ethane
Propane
Butane
Pentane
Formula
CH4
C2 H6
C3 H8
C4H10
C5H12
Molar Mass
16
30
44
58
72
Density (g/cm3)
.00065
1.28
.493
.00248
.626
18. Between propane and butane, which has the higher boiling point? Choice
the correct answer with the correct reason.
A. Propane, because it is denser
B. Propane, because it has stronger bonds between the carbon and
hydrogen
C. Butane, because it has more polar characteristic
D. Butane, because it is heavier (more polarizable)
19. A student burns butane and collects the resulting carbon dioxide gas
over water at room temperature. She then calculates the density using the
ideal gas law equation. She makes one mistake in the calculation and
obtains a density that is very slightly different than the actual density for
carbon dioxide. What was most likely her mistake and what was her
calculated density in relationship to the actual density? Assume she made
no mistake in laboratory data collection and only made a mistake in the
calculations.
A. She forgot to subtract the vapor pressure of the water, resulting
in a lower calculated density
B. She forgot to subtract the vapor pressure of the water, resulting
in a higher calculated density
C. She forgot to convert the temperature to Kelvins, resulting in a
lower calculated density
D. She forgot to convert the temperature to Kelvins, resulting in a
higher calculated density
20. The student calculates the root mean square speed of butane and one
of the other gases. She finds that the other molecules of the other gas
travel an average of almost twice the speed of butane. Assume they are
both ideal gasses. Which gas is the unknown? (SAME TEMPERATURE)
A. Methane
B. Ethane
C. Propane
D. Pentane
21)
Polymers are chains of hydrocarbons that are often strengthened by hydrogen
bonds between the chains. Which of the following correctly show the use of
Hydrogen bonds between the polymer chains?
A)
B)
C)
D)
22)
The following picture depicts the general area where fish swim. Fish use oxygen that is dissolved in the water to breath. Based On this statement and known solubility rules, which month would have the most O2 dissolved in the water? Assume the lake is in Maine where lakes begin to freeze in early December
a) July
b) October
c) January
d) There would be the same amount year round
23). The heating curve of an unknown substance is given. Which
substance does this heating curve represent?
a) Ethanol
b) Water
c) Iron
d) Sodium
The following questions (24-26) apply to this chart, one point will be given for
completing the chart correctly.
Compound
IMF
Sulfate
Sodium Chloride
Iodine
Carbon Monoxide
Rubidium Floride
Mass
Polarity
Estimated BP
24.) Which of the elements (substances) in the table would be expected to have a higher
boiling point than KCl?
A.) Sodium Chloride B.) Carbon Monoxide
C.) Iodine
D.) Rubidium Floride
25.) If you were to place the compounds given in a chromatagram (chromatography?) of
non-polar siding, which would be expected to elute less rapidly than Francium Oxide but
more rapidly than Barium Nitride?
A.) Iodine B.) Sodium Chloride C.) Rubidium Floride
D.) Sulfate
26.) Which of the non-polar compounds would have the highest boiling point and for
what reason?
A.) Sulfate, due to the strength of its Intermolecular force
B.) Sodium Chloride, due to its molar mass
C.) Iodine, due to its molar mass
D.) Carbon monoxide, due to its density
27. A manometer containing water and a manometer containing mercury are in the
same room with constant temperature and pressure. In which manometer will the
difference in height of the liquid in the tubes be lowest?
A. The manometer containing water, because water is a polar molecule
with a high dipole-dipole moment and is capable of forming hydrogen
bonds, and these forces of attraction between molecules enable the water
to be adhesive to the manometer, and resist the pressures of the gas and
the atmosphere.
B. The manometer containing water, because water is less dense than
mercury, and therefore requires more pressure to move so it will move less
within the manometer than the mercury will. c
C. The manometer containing mercury, because mercury is a metal,
and metals have a high malleability which allows for the mercury to bend
slightly under the pressure while maintaining its position, thus decreasing
its movement within the manometer.
D. The manometer containing mercury, because mercury is denser than
water, and therefore requires more pressure to move so it will move less
within the manometer than the mercury will, as both are under the same
pressure,
28. A student observes one manometer containing mercury, and three manometers
containing unknown liquids. The student records the data below. Each manometer
contains 30.0g of chlorine gas at 0˚C and 730 mm Hg. A barometer in the room reads at
0.76 m Hg.
Manometer
Mercury
Unknown liquid 1
Unknown liquid 2
Unknown liquid 3
Height of liquid in first tube
60.0 mm
816 mm
4.40 mm
25.0 mm
Height of liquid in se
30.0 mm
408 mm
2.20 mm
38.6 mm
Which unknown liquid contains water? Note: mercury is 13.6 times denser than water.
A. Unknown liquid 1
B.
Unknown liquid 2
C.
Unknown liquid 3
D. None of the manometers contain water
29. Energy is required to change liquid water to water vapor because
A. The bonds between atoms in the H2O (l) molecules require energy to
break and become H2O (g).
B. The intermolecular forces, such as hydrogen bonds, between the H2O
(l) molecules require energy to be overcome, and the molecules can only
separate to become a gas once these intermolecular forces are overcome.
C. Water vapor is hotter than or equal in temperature to water at
constant pressure, and therefore kinetic energy is required to create this
change in temperature, which results in a change in state from liquid to
gas.
D.
All of the above.
30. 60.0 g of carbon tetrachloride is heated from solid at -52°C to gas at 99°C, with a
liquid phase in between. The substance has a melting point at -22.92°C, a boiling point at
76.72°C, and a specific heat of 0.866 J/g*°C. It’s enthalpy of fusion is 2.67 kJ/mol, and
its enthalpy of vaporization is 30.0 kJ/mol. How much heat needs to be added to
complete the process?
(A) 1.27 * 10 J
(B) 7.85 * 10 J
(C) 2.06 * 10 J
(D) 2.06 * 10 J
4
3
4
3
31.
The table below shows the solubilities of four unknown compounds in silver nitrate
(AgNO ) and CuCl .
3
2
AgNO
Unknown #1
3
soluble
CuCl
2
insoluble
Unknown #2
soluble
moderately soluble
Unknown #3
insoluble
soluble
Unknown #4
moderately soluble
insoluble
Based on the information in the table, which of the following statements is true?
(A) Unknown #1 is insoluble in ammonia.
(B) Unknown #3 is soluble in ammonia but insoluble in propane.
(C) Unknown #4 is soluble in propane but insoluble in ammonia.
(D) Unknown #3 is insoluble in ammonia but soluble in propane.
32.
Which of the following is most likely the identity of the unknown substance, based on the
information in the table below?
Structure
Linear
Type of Bonding
Nonpolar covalent
IMF Type
London dispersion forces
IMF Strength
Strong
(A) CH
(B) CI
(C) Kr
(D) SeI
4
4
2
Use the following information for questions 33 and 34.
A student was given 4 different gases at 1 atm and 273 K, and was told that he had
one each of sulfur dioxide, propane, decane, and propanoic acid. The gases were
then condensed, made into four 250 mL solutions, and run through a nonpolar
column for a chromatography test. The data are shown below.
Gaseous State
Compound
Volume (L)
D
15.68
B
8.960
C
8.960
A
19.04
Chromatography Readout
33. What are the identities of the compounds in the order of A, B, C, D?
(respectively)
A) sulfur dioxide, propanoic acid, propane, decane
B) sulfur dioxide, propanoic acid, decane, propane
C) propane, decane, propanoic acid, sulfur dioxide
D) propane, sulfur dioxide, propanoic acid, decane
34. Assuming compound C is SO , what would the speed of sound in SO gas be
if 245.8 joules were added to the gas before the condensation and
chromatography? Assume that the pressure remained constant at 1 atm.
2
2
Specific Heat of SO : 0.64 J/g K
Speed of Sound in SO : 1.29*R*TM
R = 8.3145 J/mol K
T = temperature in K
M = molar mass in kg/mol
2
2
A) 6.947 m/s
B) 216.2 m/s
C) 6.838 m/s
D) 219.7 m/s
35. The rms speed of ammonia is greater than that of an unknown compound at the
same temperature, while the boiling point of the unknown is higher. Which of the
following could be the unknown?
A) NaCl
B) H O
C) H
D) C H
2
2
2
6
Use the following data set to answer questions 36-38. A scientist is observing different
compounds in the laboratory. The data the scientist collects is shown below:
Compound
Boiling
Point
Water Solubility
(0ºC)
Conductivity in
Water
Attraction to
Magnetic Field
A
-33ºC
Insoluble
Low
None
B
69ºC
Insoluble
Slight
None
C
158ºC
Very Slightly
Soluble
Slight
Slight
D
-1ºC
Insoluble
Slight
None
36, Using the data above, which compound contains the strongest IMFs?
A. D
B. C
C. A
D. B
37. The scientist identifies compound D as butane. If the scientist isolates butane
in a rigid container and increases pressure inside the container, which of the
following scenarios are possible?
I. The solubility in water increases
II. The boiling point will rise
III. The gas solidifies
A. I and II
B. I and III
C. II and III
D. II only
38. The scientist made an error in the lab data. If LiNO3 can dissolve in the liquid
form of A, which piece of data is most probably incorrect?
A. Boiling Point
B. Attraction to Magnetic Field
C. Conductivity in Water
D. Water Solubility
39. 37.0 grams of 1-propanol is dissolved into 60.0 grams of distilled water. What is the
mole fraction of 1-propanol?
A. 0.233
B. 0.616
C. 0.156
D. 0.185
40. The density of gas A is found to be 54 g/L. It occupies 34.5 L at 400.0 torr and 42 ºC.
What is its molar mass?
A. 3.4 g/mol
B. 2600 g/mol
C. 8.4 g/mol
D. 350 g/mol
41. Order the compounds below in terms of polarity, from least polar to most polar. Then
predict which would elute furthest in a chromatogram. (What type of chromatography?)
Methanol, Heptanol, 1-butanol
A. Methanol, 1-butanol, Heptanol; 1-butanol
B. Heptanol, 1-butanol, methanol; Heptanol
C. 1-butanol, Heptanol, methanol; Methanol
D. Heptanol, methanol, 1-butanol; Heptanol
42. An unknown gas has a higher boiling point greater than Cl2 and a lower
boiling point than methanol (CH4O). Choose the response that best fits the
identity of the gas and the reasoning behind it.
A. F2, because the LDFs are lower in this molecule than in chlorine gas, and it is
not polar
B. F2, because the LDFs are higher in the molecule than in chlorine gas, and it is
not polar
C. Diethyl ether (C4H10O), because it more polar than Cl2 and will need energy to
boil and break the bonds in the molecule
D. Diethyl ether (C4H10O), because it is less polar than ethanol it will not need as
much energy to become a gas
43. There is a mixture of helium and hydrogen gases and there are exactly twice
as many moles of helium gas as hydrogen gas in the container that contains 10.
g of gas total. What is the density of the mixture of gases at 100K and 1.5 atm
(g/L)?
Gas
Density
Hydrogen Gas
0.000089 g/mL
Helium Gas
0.1786 g/L
A. 0.608 g/L
B. 0.149 g/L
C. 0.134 g/L
D. 0.548 g/L
44. The flash point of a compound is the minimum temperature at which
enough vapors are given off to ignite in air. Compounds with a flash point
above 37.8 degrees Celsius are considered combustible, while those
below 37.8 degrees Celsius are considered flammable. The vapor
pressure of a compound indicates a compounds evaporation rate. Which
of the following compounds is combustible?
A. C3H3NO
B. C3H7Cl
C. C3H6F2
D. C3H6S
45. The log Koc value is commonly used to determine the adsorption of
chemicals by different soils, where a higher log Koc value leads to more
adsorption onto soil and the possibility of contaminating surface water.
The equations log Kd = a*log Kow + log Foc + b (for constant positive a
and b) and log Kd - log Foc = log (Koc/100) relate the log Koc value to the
log Kow (calculated as log( [solute in octanol] / [solute in water] )). Which
of the following compounds would have the lowest log Koc value?
A. C3H3NO
B. C3H7Cl.
C. C3H6F2
E. C3H6S
The following measurements are taken when two gases are added to a sealed
rigid container. Use this table to answer questions 46.
Time
Temperature
Pressure
0
26
1.12
5
27
1.00
10
26.5
0.97
46. Which is the most likely reaction occurring in the container?
A. H2 + Cl2 → 2 HCl
B. CO2 + 2 Cl2 → CCl4 + O2
C. 2 NH3 + 2 O2 → 2 NO2 + 3 H2
D. SCl2 + F2 → SF2 + Cl2
47. Three unknown compounds are boiled and collected in uninflated balloons at
atmospheric pressure. Each balloon and maintained at 90 ˚C and tied off and
massed when it deflates to a volume of 600 mL. The same mass of each
compound is boiled again, this time measuring the energy added to completely
vaporize it. The following measurements are taken.
Compound
Mass
Energy absorbed
X
5.0 g
500 J
Y
2.8 g
200 J
Z
0.3 g
278 J
Which compound condenses first?
A. Compound X
B. Compound Y
C. Compound Z
D. Cannot be determined
48. Which liquid absorbs the most energy when GeO2 is dissolved in it?
A. Compound X
B. Compound Y
C. Compound Z
D. Cannot be determined
49.
Molecular Weight (g/mol) Dipole Moment u(D)
3 -­‐ Octyne
110
1.12
Ammonia
17
1.42
Potassium Iodide
166
10.8
Xenon Hexafluoride
245
0.00
Which of the above compounds will boil at the highest temperature?
A. Potassium Iodide B. Xenon Hexafluoride C. 3-­‐Octyne D. Ammonia 50. 2C8H14(g) + 46O2(g) → 16CO2(g) + 14H2O(g)
The above reaction takes place at 1 atm, 133°C and with 22.50L of 3-­‐octyne with excess oxygen. How many liters of water are produced?
A. 22.50L B. 314.89L C. 315.00L D. 314.83L 51. How many peaks would a chromatography graph of MgCl2 have? (The cylinder is lined with a polar substance)
A. 0
B. 1
C. 2
D. 3
52. Consider the following graph. The four curves each represent one of the four
compounds: C4H10O, CH3OH, H2
O or C6H5NH2.
Which of the following correctly identifies the curves with the compounds?
A. 1 = LiCl, 2 = CH3OH, 3 = H2O, 4 = C4H10O
B. 1 = LiCl, 2 = H2O, 3 = CH3OH, 4 = C4H10O
C. 1 = C4H10O, 2 = CH3OH, 3 = H2O, 4 = LiCl
D. 1 = C4H10O, 2 = H2O, 3 = CH3OH, 4 = LiCl
53. Which of the following ranks the four compounds correctly from lowest boiling point to
highest boiling point?
A. LiCl < H2O < CH3OH < C4H10O
B. C4H10O < CH3OH < H2O < LiCl
C. LiCl < CH3OH < H2O < C4H10O
D. C4H10O < H2O < CH3OH < LiCl
54. A solution of methanol has a vapor pressure that is at regular atmospheric pressure. How
much heat is required to heat 10.00 g of methanol from this liquid temperature to a 80.00ºC gas?
The specific heat of methanol is 2.510 J/gºC. The enthalpy of fusion is 0.5900 kJ/mol. The
enthalpy of vaporization is 35.21 kJ/mol. The boiling point of this compound is 64.70°C.
A. 384.0 ºC
B. 736.1 ºC
C. 395.0 ºC
D. 11.00 ºC
A solution is analyzed via column chromatography (nonpolar column). The following
questions are based on the resulting chromatogram. (assume the peaks are numbered
in order)
55. Which of the substances was the most nonpolar?
a. A. Peak 1
b. B. Peak 2
c. C. Peak 3
d. D. None of the above, they are all polar
e.
56. Which substance exhibits the most dipole-dipole forces?
a. Peak 1
b. Peak 2
c. Peak 3
d. None of the above, they are all nonpolar
57.
Which of the substances has the lowest boiling point?
a. Peak 1
b. Peak 2
c. Peak 3
d. Cannot be determined from the information given. A chromatogram has no
information from which boiling point can be determined.
ANSWERS 1. C
combined gas law, P1T1 = P2T2
Students would want to say that the pressure would triple, but the temperature must
change to Kelvin. The temperature thus, would be increased by a factor less than three.
573/373 < 3 Correct and good question – I think many students would make that mistake
2. D
Less NRG is required to dissolve the CaCl2 than is released. In exothermic reactions,
energy is released, in this case, by the way of heat. Weak bonds have been broken with
calcium chloride. Clearly not B or C – how does on determine if the bonds are weak? A
may be a good answer too – good use of MW in data table. 3. The answer is C. oxygen. When calculating the rms value
(rms=(3RT/M)1/2) of neon gas under these conditions, the answer is
rms=461.28 J. After substituting that value into another rms equation
for the unknown gas, the molar mass of the gas is found to be 32
g/mol. The identity of this gas is oxygen because it is diatomic and its
atomic mass is 16 g. They should have the same molar mass if same
rate and temp (20 g/mol)
4. The answer to this question is D. The density of neon would be greater
than that of the unknown. This can be found using the PM=DRT
equation. Assuming the student calculated the correct molar mass of
the unknown in the previous equation, the density of the unknown
should be 1.428 g/L and the density of neon should be 2.157 g/L. The
density of neon is greater but is not two times the density of the
unknown. Best answer is A – based on #3
5. The answer should be A. solution 1. Butanol is polar so it would be
soluble in water. It also would have a medium boiling point because of
its polar nature, dipole dipole forces, and a hydrogen bond. Its boiling
point would not be extreme either way because it would not be high,
like an ionic compound, or very low, like a nonpolar compound or a
polar compound without a hydrogen bond. The other solutions would
most likely fall into one of those categories based on their boiling
points. Correct
6. C is the correct answer to question one because a crystal lattice is
much stronger than any intermolecular forces between two covalent
molecules. This means hydrogen chloride is wrong eliminating choices
a and b, and because potassium carbonate is an ionic compound, it
does not have London dispersion forces.
CORRECT
7. B is the correct answer to question two because it is the only one that
calculates the volume of the CO2 gas correctly. A is wrong because it
does not subtract the vapor pressure of the water that the CO2 is being
collected over. C is wrong because it wrongly identifies potassium
carbonate as KCO3 instead of K2CO3 reducing its molar mass by 39.
And d is wrong because it makes both of the previously made
mistakes. One way to correctly solve this problem is by calculating the
number of moles as so:
2g K2CO3
1 mol K2CO3
1 mol CO2
138g K2CO3
1 mol K2CO3
=.014 mol CO2
And then put that into the ideal gas law equation with the pressure that
has been modified for the water vapor:
PV=nRT
746.2V=.014(62.36)298
V=.349L
V=349mL
CORRECT
8. C is the correct answer for three because it is the only one that
correctly calculates the volume of concentrated hydrogen chloride
needed to complete the reaction. A is wrong because it uses and
unbalanced equation. B is wrong because it misidentifies potassium
carbonate as KCO3 and uses an unbalanced reaction. And d is wrong
because it misidentifies potassium carbonate as KCO3. One way to
correctly calculate the volume of hydrochloric acid needed is to first
calculate the number of moles of hydrochloric acid needed as so:
2g K2CO3
1 mol K2CO3
2 mol HCl
138g K2CO3
1 mol K2CO3
.03mol HCl needed
And then use a ratio to solve for the volume:
18 mol
.03 mol
1L
x
x= .00167L
1.67mL CORRECT
9. C. 5.7 * 10^−1 L
Given that each nanoparticle has an average volume of 5.236*10^-1 µm^3, using the
density and some unit conversion, the mass of each nanoparticle can be determined:
5.236*10^-1 µm^3 * (1 cm/10^4 µm)^3 * 2.65 g/cm^3 = 1.388*10^-12 g
Given that the number of nanoparticles is given (albeit in moles), the total mass of SiO2
can be determined:
1.388*10^-12 g/particle * 3.0 * 10^-12 moles particles * (6.022*10^23/mole) =
2.508 g
Given this mass, then number of moles of SiO2 can be determined using its molar mass:
2.508 g SiO2 * (1 mol SiO2/60.08 g SiO2) = .0417 mol SiO2
By balancing the reaction described, we determine:
SiC8H20O4 + 2 H2O -> SiO2 + 4 C2H5OH
Using stoichiometry, the moles of tetraethyl orthosilicate can be determined:
.0417 mol SiO2 * (1 mol SiC8H20O4/1 mol SiO2) = .0417 mol SiC8H20O4
Using the ideal gas law with the conditions provided, we can determine:
PV = nRT
(3 atm)V = (.0417 mol)(.08206 L atm/(K mol))(500 K)
V = .57 L = 5.7*10^-1 L CORRECT (and very complicated)
10. B. I and III
We examine each answer choice individually:
I. Ethane would not dissolve readily in the solution
This answer choice is true, as ethane is a non-polar hydrocarbon, whereas both
water and ethanol are polar compounds. As such, there would be minimal IMFs between
ethane and any constituent of the solution, causing it not to dissolve. In other words,
because water is polar and ethane is non-polar and “like dissolves like”, the two unlike
compounds will not dissolve.
II. Whether ethanol is fully dissolved is dependent on the volume of water initially in the
tank
This answer choice is not true, as ethanol is fully miscible in water, meaning that
it dissolves in any proportion. This piece of information, even if not previously known,
was given in the problem declaration. This choice tests whether the concept of miscibility
is understood.
III. Ethanol’s molarity cannot exceed 17.13 M
This answer choice is true. To come to this conclusion, we examine the solution
as the concentration of ethanol increases: the molarity increases as the solution
approaches pure ethanol. Given that the initial volume of water is unknown, we can
determine the molarity of pure ethanol to determine the maximum molarity of the final
solution (and thereby verify the upper bound given). Given that molarity is defined as the
number of moles of solute for some given volume, we can apply the density given to
deduce the maximum molarity. We do so by calculating the volume of 1 mole of ethanol:
1 mol C2H5OH * (46.07 g C2H5OH/1 mol C2H5OH) = 46.07 g C2H5OH
46.07 g C2H5OH * (1 cm^3/.789 g) = 58.39 cm^3 C2H5OH = 58.39 mL C2H5OH
58.39 mL C2H5OH * (1 L / 1000 mL) = .05839 L C2H5OH
M = mol / L
M = 1 mol/.05839L
M = 17.12M
As such, the molarity approaches 17.12M as the solution approaches pure ethanol. Therefore, the molarity of ethanol can never exceed 17.13, making choice C true. Correct (and v complicated) 11. D. Ethanol, propanol, water This question tests understanding of IMFs and their role in determining boiling point. To come up with this answer, we first determine the difference in IMFs between water and the two alcohols. Given that both alcohols only have one O-­‐H bond where hydrogen bonding can occur whereas water has two, water is likely to have more substantial hydrogen bonding. This results in greater IMFs for water than for both ethanol and propanol. Given that greater IMFs entail larger amounts of energy needed to overcome them (and change the substance’s state from a liquid to a gas), water should have the highest boiling point. Given that ethanol and propanol will both have comparable amounts of hydrogen bonding, London dispersion forces become the determining factor. The molar mass of propanol is greater than that of ethanol, giving it more substantial London dispersion forces and therefore greater IMFs. This results in propanol having a higher boiling point than ethanol, leading to the final ordering of ethanol, propanol, water. CORRECT 12. A. X has the lowest melting point due to the weakest amount of IMF between molecules. Since it is non-­‐polar, the IMF type in this compound would just be LDF. As a result, compound Y would have a higher melting point since it is polar covalent and has the same molar mass as compound X. The polar covalent bonds cause this compound to have dipole-­‐dipole interactions between molecules making it harder to overcome the IMF than the IMF in compound X. Compound Z would have a higher melting point than Y due to its high molar mass. Although it is non-­‐polar and only has LDF interactions between molecules, the large molar mass causes the LDF to become stronger. As a result, the IMF in Z requires more energy to overcome than the IMF in Y. W has the highest melting point because it is ionic. CORRECT 13. C. W and Y would exit the column first because they have ionic and polar covalent bonds respectively. The two compounds would not be attracted to the non-­‐polar characteristic of the lining, whereas the non-­‐polar compounds, X and Z, would be. Therefore X and Z would elute much slower. CORRECT 14. D. 2 C2H6 + 7O2 à 4CO-­‐2 + 6H2O 5.00 g C2H6 x (1mol / 30.068g) x ( 4 CO-­‐2 / 2 C2H6) = .322 mol CO-­‐2 Pt = PH20 + PCO2 3.00 atm = (187.5/760 atm) + PCO2 PCO2 = 2.75 atm PV = nRT V = ((.322 mol)(.08206 L atm/mol K)(273 + 65 K))/(2.75 atm) V = 3.25 L CORRECT 15. The correct answer is C because SrCl2 is an ionic compound, eliminating B. Because of the sharing of electrons, it has the greatest boiling point out of all four options. This is a distractor because people might now see the word "second" in the question and just choose the compound with the highest boiling point. Also, H2O exhibits hydrogen bonding, and so does 2-­‐heptanol. Because 2-­‐heptanol has a greater atomic mass than H2O, it has greater London dispersion and therefore, has a higher boiling point. People might choose H2O because they might not realize 2-­‐heptanol has a hydrogen bond; students might believe it only has London dispersion forces. Benzene is not a reasonable choice because it is not an ionic compound and does not have a hydrogen bond. Also, it is non polar; therefore, it only has London dispersion forces. 16. The correct answer is D because the central atoms there can be different locations of the double bonds in each carbon-­‐carbon bond. Some kids might realize the location of the pi bonds can change, so they might not pick the correct answer. H2O and SrCl2 do not make sense because SrCl2 is a ionic compound and H2O is a bent molecule and the hydrogens can only have 2 electrons in their outermost shell. H cannot be the central atom. 2-­‐heptanol is a distractor, and students might pick this answer because they might believe if they draw out the molecular structure, the alcohol can go on the left side. However, if they draw it a second time, they might put it on the right side. Unfortunately, this is not resonance because it is the same structure. 17. The answer is D because bonds are not broken during a phase change. A, B, and C are all distractors because a student might think that bonds in a covalent compound don't break and only ionic compounds' bonds do, meaning they would pick A. A student might pick B because it is the highest temperature, meaning the bonds are the strongest. Finally, a student might choose C because if they realize bonds in a ionic or covalent compounds, don't break, they might be wondering what type of bonding is in an amorphous solid. Because they have ruled out A and B, they could pick C. 18. Choice D is the correct answer. Both butane and propane have limited polar characteristics. As such, their relative boiling points are mostly determined on which molecule has more London Dispersion Forces, which is based off of molar mass. The correct answer would therefore be D, as it gives this reasoning along with the correct answer that butane has a higher boiling point, as it is heavier. CORRECT 19. Choice B is the best answer choice. The two possible choice of mistakes are that she forgot to subtract vapor pressure, resulting in a higher pressure, or she forgot to convert temperature to Kelvins, using a lower temperature. Using the alternate form of the ideal gas law equation, PM=DRT, it should be realized that, assuming everything else is constant, a higher pressure will yield a higher density, while a lower temperature will also yield a higher calculated density. Therefore, one can eliminate choices A and C. At room temperature, the negligibility of converting to Kelvins would result in a hugely different calculated density. In the original problem, it is stated the difference is slight. Therefore, the best choice would be B, as forgetting to subtract vapor pressure of water at room temperature would result in a slight difference in calculated density. CORRECT – not sure about the difference between choices B and D being apparent. 20.
The correct choice is choice A. A rms that is almost two times as fast would mean a molar mass that is nearly 4 times as small. The substance with a molar mass nearly one fourth of the molar mass of butane is methane. CORRECT – can only be done if the 2 gases are at the same temperature. 21. Reason: The answer is letter A because the said Hydrogen bonds are between Hydrogen and Oxygen. This is a valid hydrogen bond because the highly electronegative O is attracted to the H bonded to another high electronegative atom in this case N. Letter C and B are distracters because they contain bonds between two hydrogen atoms. If a student were to not fully understand hydrogen bonds, they might think that hydrogen bonds are between two hydrogen atoms. Both C and D have a double bond on hydrogen which isn’t possible so those two could be ruled out, but without fully understanding Hydrogen bonds a student might pick B. CORRECT 22. The answer is B because the water has had time to get cold after the summer, and gases are more soluble in the cold. Both A and C are misleading because some might think the warmer the lake the more O2 can be dissolved because that is the case with solids. Others might look for the coldest month (January), but the given info says the lake freezes which would not allow oxygen to be dissolved into the lake because it is sealed off. CORRECT – although not sure about C-­‐ would oxygen already be dissolved, or permeate into the system through the ice? 23. The answer is C Iron. A student might be confused by the two metals, but since Iron has more electrons there is a greater attraction, causing the iron to have a larger boiling point. Since around 2,800°C is a large boil point the student should be able to see that the heating curve represents iron. CORRECT 24. Answer: D, because it would have both a stronger intermolecular force and also a larger molar mass than KCl, however Sodium Chloride would not although it would be the next closest option Better answer is NaCl – Na is smaller than K and due to Coulumb’s law will have greater lattice NRG – looking these up online, one finds that RbF and NaCl are very close in bp 25. Answer: B, because it would be more polar than Barium Nitride, but less polar than Francium Oxide therefore cause it to elute at a higher rate than Barium Nitride but not the Francium Oxide due to the non-­‐polar lining of the chromatagram. Option "C" is meant to be an attractor (distractor?), but it incorrect because it would elute more rapidly than both compounds named in the problem. Hard to tell without eneg values – also sulfate is an ion and not a compound 26. Answer: C, because both Iodine and Sulfate are non-­‐polar, the only non-­‐polar compounds of the group, but Iodine has a higher molar mass so therefore it would have a higher boiling point than Sulfate, Sulfate is meant to be an (distractor) attractor since it is the only other non-­‐polar compound. CORRECT – although sulfate is not a compound – also the reason would be better stated as more LDF or being more polarizable 27. Answer: D. Rationale: Pressure is what causes the liquid within a manometer to move. Pressure is defined as force/area. Force is defined as mass times acceleration. Density is defined as mass/volume, and because volume between the manometers is constant and because mercury is denser than water, there is more mass of liquid in the mercury manometer than in the water manometer. Acceleration is also constant (assuming this is done on earth, the acceleration is gravity, or 9.8 m/s2). Thus, because force is mass times acceleration, and the mercury manometer has more mass of liquid in the same volume of liquid, the force required to move the mercury will be greater than the force required to push the water. Pressure is defined as force/area, and area is constant between the manometers, so the mercury will require more pressure to move it than the water will, resulting in the same pressure from the gas moving the water more than the mercury, causing a lower difference in height in the tubes of mercury manometer. All incorrect answers state some correct information, but are incorrect because the information stated is less or not at all relevant to the topic in question than the information stated in answer D. CORRECT 28. Answer: A. Rationale: Water is 13.6 times less dense than mercury, and therefore the height difference of the water in the manometer will be 13.6 times greater than the mercury in the other manometer. 408 mm is the height difference of the liquids in the manometer containing unknown liquid 1, which is 13.6 times 30 mm, which is the height difference of the liquids in the manometer containing mercury. B is incorrect because the height difference in the manometer containing unknown liquid 2 is 2.2 mm, which is 30 mm divided by 13.6, indicating that unknown liquid 2 is actually 13.6 times denser than mercury, not less dense. C is incorrect because the height difference in the manometer containing unknown liquid 3 is 13.6, which is not 13.6 times greater than the height difference in the manometer containing mercury. CORRECT 29. Answer: B Rationale: During a phase change, the intermolecular forces between particles are weakened or broken. An ideal gas has no attraction between its particles, and this is because there is enough kinetic energy in the gas to overcome the intermolecular forces of attraction between the particles, thus removing all attraction among the particles. The answer is not A because during a phase change, the intramolecular forces of attraction are not broken. The answer is not C because, although water vapor is hotter than or equal in temperature to water at constant pressure, the temperature does not cause the phase change; it is merely a side effect of the kinetic energy necessary to bring about the phase change being present. CORRECT 30. (C) The first step in this process is to heat up the carbon tetrachloride from -­‐52°C to -­‐22.92°C, its boiling (melting) point. The amount of heat needed to do this would be determined by multiplying the mass of the solid, 60.0 g, by its specific heat, 0.866 J/g*°C, multiplied by the change in temperature, which in this step is (-­‐22.92-­‐(-­‐52)) or 29.08°C. This yields a result of 1510.99 J. The next step would be to add the required heat of fusion to melt the solid with no change in temperature. The amount of heat needed to do this would be determined by multiplying the number of moles by the enthalpy of fusion. The number of moles of CCl4 present is 60.0 g / (153.8 g/mol) = 0.390 mol. Thus, 0.390 mol is multiplied by the enthalpy of fusion, 2670 J/mol, to get 1041.5 J. Next, we would need to heat up the resulting liquid to its boiling point, 76.72°C. The amount of heat required in this step would be 60.0 g* (0.866 J/g*°C) * (76.72-­‐(-­‐
22.92)°C = 5177.3 J. After this is complete, the heat of vaporization would need to be added to change the liquid to a gas. The amount of heat needed to do this would be determined by 0.390 mol * 30,000 J/mol = 11701.8 J. Finally, the resulting gas would need to be heated up to 99°C. The amount of heat needed in this step is 60.0 g *(0.866 J/g*°C) *(99-­‐76.72)°C = 1157.67 J. The total amount of heat required for the process is found by adding the amount of heat, in joules, needed for each step. Adding 1510.99 + 1041.5 + 5177.3 + 11071.8 + 1157.67 = 2.06 * 104 J. CORRECT 31. (D) There is a general chemistry rule that says “like dissolves like”, meaning that polar substances dissolve in polar substances, and nonpolar substances dissolve in nonpolar substances. Based on Lewis Structures, hybridizations and shapes, AgNO3 and ammonia are polar, and CuCl2 and propane are nonpolar. Choice D is the only one that correctly matches the insolubility in AgNO3 with the insolubility in ammonia, as well as the solubility in CuCl2 with the solubility in propane. CORRECT 32. (D) All four answer choices have nonpolar covalent bonding, but CH4 and CI4 have a tetrahedral, not linear, structure. Thus, choices A and B can be eliminated. Kr and SeI2 both have London Dispersion Forces, but the table says that the LDF’s are strong. The molar mass of SeI2 is significantly larger than that of Kr, so it can be determined that SeI2 has the stronger intermolecular forces. Therefore, answer choice D is the most logical. CORRECT -­‐ note that Kr does not have a shape (its only a atom) 33. A) Correct answer: Sulfur dioxide is polar and doesn’t interact much with the nonpolar stationary phase, so it passes through fairly quickly. Propanoic acid is also a polar molecule, so it has the next lowest time. Between propane and decane, decane has a higher molecular mass and therefore more London Dispersion Forces, so it slightly more time to pass through than propane does. B) This would be chosen if the student either forgot about London Dispersion Forces, or thought that they worked in reverse. C) A student would pick this if they had nonpolar and polar switched, mistakenly believing that polar molecules and nonpolar molecules interact more. D) If the student had only based their answer off of London Dispersion Forces, they would choose this answer. CORRECT 34. A) Performing calculations with M in g/mol instead of kg/mol would yield this result. B) Solving the problem without calculating the number of moles from the volume given at STP (using 1 mole instead of the correct 0.4 moles) would give this answer. C) If the student committed both of the above errors, they would arrive at this response. D) Correct answer: Moles of SO2: 8.96 L22.4 L/mol=0.4 mol SO2 (the gas is already at STP) Change in temperature: q = msT T=qms=245.8 J(0.4 mol)*(64 g/mol)*(0.64 J/g K)=15 K New volume: V1T1=V2T2 (pressure remained constant) V2=V1T2T1=(8.96 L)*(288 K)273 K=9.452 L (is this needed?) Speed of Sound in SO2: 1.29*(8.3145 J/mol K)*(288 K)0.064 kg/mol=219.7 m/s (CORRECT) 35. A) Forgetting about the importance of dipole moment in determining relative boiling point would lead to this answer. (this is also an ionic compound and has a very high bp) B) Correct answer: Water has a higher molar mass, and therefore a lower rms speed, than that of ammonia. It also has hydrogen bonds that raise its boiling point higher than that of ammonia. C) Incorrectly believing that a higher molar mass corresponds to a higher rms speed would result in this answer. D) C2H6 has a higher molar mass than ammonia, so it fits the rms speed condition, but it is nonpolar, so someone who didn’t recognize this fact might pick this choice’ 36. The answer is choice B; the compound with the highest IMF will have the highest boiling point. Compound C, choice B, is the one with clearly the highest boiling point and therefore highest IMF. CORRECT 37. The answer is choice is A. The solubility of a gas in water increases as pressure increases. The boiling point of a gas increases as pressure increases because more downward force is containing the liquid form of the compound from boiling into a gas. B and C are incorrect because statement III states that the gas will solidify. However a gas is more prone to liquify as pressure increases. CORRECT -­‐ although with a high enough pressure, it could solidify 38. The answer choice is D. If an ionic compound dissolves in a liquid, the liquid is most likely to be polar. If the compound is polar, it should dissolve in water. However, the scientist incorrectly observed water solubility. An argument for choice A could be made, but water solubility is a greater indicator of polarity. A polar covalent compound is not conductive in water, and the attraction to a magnetic field says nothing about polarity; one has to know the molecular geometry and see if there are unbonded electrons on the central atom. CORRECT 39. (C) First, the molecular formula for 1-­‐propanol is C3H8O. The number of moles of 37 g of it was found using stoichiometry: 37 g x 1 mol / 60.03 g = 37/60.03 = 0.616 mol. The moles of water involved was found: 60 g x 1 mol/18.02 g = 60/18.02 = 3.33 mol. Mole fraction is found by (moles of solute) / (total moles). Total moles = (3.33 + 0.616) = 3.946 mol. The fraction would be 0.616 / 3.946 = 0.156. CORRECT 40: (B) Use the version of the ideal gas law, PM = DRT, this problem can be solved. P = 400/760, D = 54 g/L, R is the gas constant 0.0821, and T is (42+273) or 315 K. (400/760)(M) = (53 g/L)(0.0821)(315) M, or molar mass, is 2604.25 or 2600 g/mol. CORRECT 41. (B) Firstly, the molecular formulas for methanol, heptanol, and 1-­‐butanol are respectively CH4O, C7H16O, and C4H10O. Carbon atoms in the compound make the compound more nonpolar. Therefore, the order would be Heptanol, 1-­‐butanol, and methanol. The chromatogram is lined with nonpolar substances. The more nonpolar of the compounds would elute further. Since heptanol is the least polar, and thus most nonpolar, it would elute furthest. The polarity order is correct, the chromatogram would be correct if the column is nonpolar. 42.The answer is D. Both A and B are eliminated from being the answer because F2 should have lower boiling point than the chlorine gas since there are a lower amount of LDFs since fluorine is a smaller atom/molecule than chlorine. Answer C is the distractor as it accounts for why the answer choice of diethyl ether is correct, but the second portion of the explanation states that the bonds within the molecule have been broken when becoming a gas. Answer D is correct because it gives the correct identity of the molecule, as well as the correct explanation. Diethyl ether does have a polar carbon-­‐
oxygen bond, meaning that it has a greater forces impacting it than LDFs in the form of dipole-­‐dipole forces, but the oxygen is not as available to be subjected to these forces as much as the oxygen in methanol because of the placement of the oxygen in the molecule. In the diethyl ether molecule, the oxygen is blocked by ethyl groups on both sides. Therefore there is less available space to for positive dipoles to bond to it. In order to be able to answer this question, the structure of alcohols and ethers must be known, as well as the effect of polarity on boiling point. CORRET – many students might pick C, however as explained here, this is incorrect 43. The correct answer is A. The number of moles total must be found. This is three as there are 2 moles of helium in the mixture and one mole of hydrogen. From there, the molar mass is found by dividing the ten grams of mixture by the number of moles (10g / 3 mol = 3.33 g/mol). This is then used in the equation PM = DRT and D is solved for(1.5 atm)(3.33g/mol) = D(0.0821)(100K). Once simplified, D comes out to be .608 g/L. Choice B is incorrect as it uses the number of liters (67.2 L) that would be present at STP in order to find the density of the mixture. Choice C is incorrect because it uses the molar mass of hydrogen given, and not that of a diatomic gas (H2), and it uses STP as the conditions that the mixture is under. Choice D is incorrect because it does not account for hydrogen being a diatomic gas. This choice is the distractor as it makes sense, if the student does not know that H2 is diatomic, they cannot solve the problem. Also, the table of densities is not needed to solve the problem. It is a distractor if they choose to solve the problem given the values in the table. CORRECT 44. From the first chart, students should realize that at a given temperature, the compound with the highest normal boiling point has the lower vapor pressure. From the second chart, students should deduce that low relative vapor pressures correspond to higher relative flash points. Thus, the compound with the highest boiling point would also have the highest flash point. Since only one answer can be correct, the student should deduce from the question that the highest flash point would correspond to the only compound that would be combustible. Thus, the question is asking for the compound with the highest boiling point. The correct answer is A, C3H3NO. This compound has C-­‐N and C-­‐O bonds, making it very polar. The compound can easily hydrogen bond, as the N and O are protruding. Because of the strong IMF, evaporating A would take more energy than the other compounds, and would have a higher boiling point. B is not a good choice, as there are no polar bonds and thus the only IMF present are London dispersion forces. C is a distraction to students. The C-­‐F bond would be extremely polar, and a hydrogen bond with Fluorine would be stronger than the hydrogen bonds with Nitrogen or Oxygen in choice A. However, there are two fluorines pulling in opposite directions, rendering the compound nonpolar as a whole. Choice D looks good to students that do not realize C-­‐S bonds are nonpolar. Otherwise, the compound would be very polar to one side, resulting in dipole-­‐dipole IMF. Following this train of thought, a student might choose D over A due to its higher molar mass, relying on the positive correlation between molar mass and boiling point to differentiate between the compounds with dipole-­‐dipole IMF. CORRECT 45. At first glance, this question appears extremely complicated. However, subtracting log Foc from both sides in the first equation yields: log Kd -­‐ log Foc = a*log Kow + b. Substituting the second equation for log Kd -­‐ log Foc yields log(Koc/100) = a*log Kow + b. Since a and b are positive constants, as log Koc increases, log Kow does as well. Students may recognize the calculation for log Kow as the same as the partition coefficient. Nevertheless, a student should realize that a nonpolar substance would have a higher concentration of solute in octanol than in water. This is because octanol is nonpolar, water is polar, and "like dissolves like". Thus, a polar substance would have a lower log Kow, and thus a lower log Koc. The question is asking for the most polar substance. Choice A is correct because the C-­‐N and C-­‐O bonds are very polar. The structure of the compound results in a dipole. Choice B is incorrect. The one C-­‐Cl bond is slightly polar, but it is completely surrounded by nonpolar bonds traveling in the same direction, so there is no way for any sort of polarity in the whole molecule. Choice C is a distraction. Students know that Fluorine is the most electronegative element, and thus C-­‐F bonds are extremely polar. However, the polarity travels in almost opposite directions so most of the polarity cancels out overall. Choice D is also a bad choice because the electronegativity difference between C and S is minuscule. To students who are unaware of this and proceed based on the "desire" of sulfur for 2 electrons, lacking knowledge of Coulomb's Law, the substance would appear polar. However, Coulomb's Law tells us that the force of attraction decreases with an increased atomic radius, justifying the closeness of the electronegativity values of Carbon and Sulfur. CORRECT 46. As pressure decreases at constant temperature and volume, molar quantity decreases according to gas laws. B is the correct option because it shows a net decrease in moles of gas within the chamber. A goes from higher to lower subscripts and shows an increase in IMFs, appearing attractive if one confuses diatomic gases for higher molar quantity or chooses deviation from the ideal gas law as the source of the decreased pressure. This is not the best answer, as B clearly shows a stronger impact on pressure by decreasing the pressure from each amount of reacting gas 50%. C shows an increase in total moles of gas and would look attractive if the wrong correlation between pressure and molar quantity was used. D follows halogen displacement, and might be selected as a guess because it is a likely reaction. It also generates strong IMFs that would look attractive if one was using the non-­‐ideal gas correction instead of the ideal relationship between pressure and molar quantity 47. A. Since compound X absorbs the most energy when boiled and heated, it has the strongest IMFs and highest boiling point. Therefore it would condense more readily and at a higher temperature. The correct answer is A. B appears attractive if low Hvap and specific heat is equated to low energy to condense the gas instead of low boiling point and weak IMFs keeping the compound volatile C appears attractive if only compound Z’s polarity is considered. The energy to mass ratio implies high energy to molar mass and thus high polar interaction, and it would be selected if LDFs are ignored D reduces the possibility of accurate guessing and appears attractive if delta H is not properly correlated to IMFs and/or boiling point CORRECT – based on PM = DRT and the fact that V,P,R and T are constant, one can equate M (molar mass) with m. 48. GeO2 is nonpolar because of its linear shape (can determine with Lewis structure or comparing to CO2). Therefore the energy released when it is dissolved in a polar solvent is very low and would create very little offset to the energy absorbed breaking IMFs of solute and solvent. Compound Z is the most polar because its high energy to mass ratio implies a high energy to molar mass by PM = DRT or equivalent calculations, implying stronger IMFs than London forces can create. Therefore C is the best answer. A appears attractive if the high energy absorbed is not compared to the high mass and its LDFs are thus mistaken for polar interactions. B appears attractive if either GeO2 is assumed to be polar because of its electronegativity difference or increased energy absorbed by solvation is confused with increased solubility instead of decreased because it appears the most nonpolar due to its energy to molar mass ratio. D reduces the possibility of accurate guessing as before and appears attractive if the energy absorption and release involved in solvation is forgotten or not understood. CORRECT 49. For the first question the correct answer is Potassium Iodide (A) because KI is an ionic lattice structure which are very polar and have high IMF. The other three are wrong because the organic compound 3-­‐octyne is very nonpolar, and has hydrogen bonding, this would be the distractor if you didn’t remember that KI was an ionic compound. Ammonia is a wrong because it is a gas at room temperature and pressure, this is because the ammonia molecule is not very polar at all, and the IMF of ammonia is just hydrogen bonds which are far weaker than ionic or covalent bonds and the ion-­‐
dipole forces. Xenon hexafluoride is a distractor because it is a noble gas that is in a compound, Xe6F is a polar molecule with an dipole -­‐ dipole for the IMF. Xenon hexafluoride sublimes at room temperature and pressure CORRECT 50. In this question the correct answer is 315.00 (C) because using stoichiometry the student would convert the number of liters of the gaseous form of 3-­‐octyne to moles using PV=nRT, then using stoichiometry convert to moles of gaseous H2O, and then to liters of H2O. Answer A is a distractor because this would catch a student who was out of time or guessing, by just putting down the liters of starting material, this is wrong because there would be no calculations done. Answer B is wrong because that would be using STP conditions for the PV=nRT equations, this is wrong because the problem specifies 133°C so using 0°C would give one the wrong answer. Answer D is wrong because this would be found by using the degrees celsius in the equations of PV=nRT, this is wrong because the gas constant is in Kelvin The answer actually appears to be 157 L of water – moles of 3-­‐octyne from gas law are 0.675, multiplying by 7 to get moles of water as 4.73 and finally converting to liters using gas law. 51. The correct answer is (C) because the compound MgCl2 has to elements that would dissociate when in solution and be drawn out when going through the tube. Answer A is wrong because there would be peaks on the graph because the ions would be drawn out by the polar lining of the cylinder as the solution is going through. Answer B is wrong because this would be if the substance was overall polar and did not dissociate in solution. Answer D is wrong and is in there as a distractor for if one just counted the number of atoms in the compound. Chromatography is usually used to separate compounds in a mixture – not ions from a chemical compound – I don’t think these would separate at all. 52. C -­‐ C4H10O has only dipole-­‐dipole attractions and Lincoln Dispersion forces, and therefore has the smallest intermolecular forces. While water and methanol both have Hydrogen bonds, dipole-­‐dipole attractions, and Lincoln Dispersion forces, methanol has one free hydrogen to bond while water has two. This causes the intermolecular forces of water to be greater. LiCl is ionic, and therefore has the strongest intermolecular forces. The top curve has the highest vapor pressure, and should correspond to the the compound with the least amount of intermolecular forces, or C4H10O. The next highest curve should be methanol, then water, and the bottom curve should be LiCl. CORRECT 53. A -­‐ C4H10O has only dipole-­‐dipole attractions and Lincoln Dispersion forces, and therefore has the smallest intermolecular forces. While water and methanol both have Hydrogen bonds, dipole-­‐dipole attractions, and Lincoln Dispersion forces, methanol has one free hydrogen to bond while water has two. This causes the intermolecular forces of water to be greater. LiCl is ionic, and therefore has the strongest intermolecular forces. As the intermolecular forces increase, the boiling point also should increase, therefore LiCl < H2O < CH3OH < C4H10O, or choice A. However this order is backwards – so B is better (lowest to highest) – also LiCl has no intermolecular forces. 54. C 10 g ·∙ = 0.3125 mol CH3OH q1 = Δ Hvap(m) = (35.21)(0.3125) = 11.00 ºC q2 = msΔT = (10.00)(2.510)(15.3) = 384.0 ºC 11.00 + 384.0 = 395.0 ºC Answer should be in J and the q1 will be in kJ based on what is given. – Answer will be different. 55. C is the correct answer A Incorrect: this took the least time to pass through the nonpolar column, meaning there was the least attraction between the column and substance and this substance is the most polar. B Incorrect: this took neither the most time nor the least time to pass through the column, thus it is neither extreme and cannot be the most polar or the most nonpolar. C Correct: this took the most time to pass through the nonpolar column, meaning that there was the greatest amount of attraction between the substance and the column, and, as like is attracted to like, the substance at peak 3 is the most nonpolar. D Incorrect: only one substance is extremely polar, another is nonpolar, and the third is in the middle of the two. CORRECT 56. A is the correct answer. f. Correct: this substance took the least time to pass through the nonpolar column. This means that there was the least amount of attraction between the column and the substance. Since like is attracted to like, this is the most polar of the substances as it exhibited the least amount of attraction between the column and the substance. Polar substances exhibit the most dipole-­‐dipole forces, so the substance at peak 1 has the most dipole-­‐dipole forces. g. Incorrect: this substance took neither the most time nor the least time to pass through the column, so it cannot be said from the given information that it is the most polar or the least polar. h. Incorrect: this substance took the most time to pass through the column, and thus is the most nonpolar. i. Incorrect: only one substance is nonpolar, another is polar, and the other is in the middle of the two. CORRECT 57.
C is the correct answer. a. Incorrect: this substance is the most polar as it passed through the column the fastest, therefore has the highest boiling point. Polar substances have greater IMF and therefore require more heat to boil. b. Incorrect: this substance is neither the most polar or the most nonpolar, so it cannot be said that this substance is either of the two extremes, so it would not have the highest or lowest boiling point. c. Correct: this substance is the most nonpolar because it took the most time to pass through the nonpolar column and nonpolar things are more attracted to other nonpolar things. Nonpolar substances have the lowest boiling point, as they have the fewest IMF. d. Incorrect: Boiling point can be ballparked from a chromatogram because boiling point increases with polarity. Actually – d is a better answer since LDF can add up quickly and increase the bp – higher than polar compounds (depending on molar mass – which is not not known)