[KEY] REVIEW – Unit 5 Test (Chp 11,13): Bonds & IMAFs in Liquids, Solids, & Solutions
Stronger intermolecular forces will cause: (write higher or lower for each)
____________ boiling point
____________ melting point
____________ viscosity
____________ heat of fusion
____________ heat of vaporization
lower
vapor pressure
(all others are higher)
1.
Based on molecular mass and dipole moment of the five compounds in the table below, which one
should have the lowest boiling point and why?
Propane b/c it’s nonpolar with the
weakest IMAFs of London
dispersion forces so it takes the least
amount of KE to separate molecules.
2.
As a liquid element freezes, the atoms get closer to/further from and they have more/less attraction
for one another. (circle each correct phrase)
3.
The weakest interparticle attractions exist between particles in the gas state and the strongest
interparticle attractions exist between particles in the solid state.
4.
Of the following molecules, which has the largest dipole moment?
(A) CO
5.
(E) F2
NH3
H2S
CH4
HCl
AlCl3
H2 O
Br2
H2 S
HCl
XeF4
Strongest IMAF (H-bonds) that
are stronger even than in NH3.
Which one of the following should have the highest boiling point?
NH3
8.
(D) HF
Which one of the following exhibits dipole-dipole attractions between molecules?
CO2
7.
(C) O2
Of the following substances, which can form hydrogen bonds?
CH3OH
6.
(B) CO2
SiH4
H2 O
dispersion forces
Of the following, __________ has the lowest heat of fusion. Weakest London
–
(smallest e cloud, least polarizable)
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9.
The predominant intermolecular force in CH3CH2NH2 is hydrogen bonding.
10.
The substance with the lowest heat of vaporization is __________.
11.
Label each of the following as an endothermic or exothermic process.
A.
B.
C.
D.
E.
melting
freezing
vaporization
sublimation
condensation
Weakest London dispersion forces
(smallest e– cloud, least polarizable)
endo
exo
endo
endo
exo
12.
A volatile liquid is one that easily vaporizes at lower temps due to weak IMAF’s.
13.
Of the following, __________ is the most volatile.
C2H6
CCl4
CCl2F2
C4H10
C6H6
Becomes gas the easiest b/c has the weakest
IMAFs (LDFs with smallest e– cloud)
14.
During the phase changes B → C and D → E…
temperature is (circle one)
increasing / decreasing / constant
kinetic energy is (circle one)
increasing / decreasing / constant
potential energy is (circle one)
increasing / decreasing / constant
Explain what the extra kinetic energy accomplishes between these points as the temperature is constant:
Added KE overcomes the IMAFs to cause more separation between molecules.
Name the phase change that is occurring between these points:
B → C melting
D → E vaporization
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15.
Molecular solids consist of atoms or molecules attracted by
A.
dipole-dipole forces
B.
London dispersion forces
C.
hydrogen bonds
D.
ANY of the above
E.
NONE of the above
16.
Amorphous solids differ from crystalline solids in that crystalline solids have organized repeated
patterns of structure. (“a” “morphous” means “without” “form”)
17.
Which statement(s) about viscosity is/are false?
(i)
(ii)
(iii)
Viscosity increases as temperature increases.
Viscosity decreases as molecular weight increases.
Viscosity decreases as intermolecular forces decrease.
More KE overcomes IMAFs and
separates molecules so (i) is false.
Higher viscosity is from stronger
IMAFs so (ii) is false.
18.
The property responsible for the "beading up" of water is surface tension and increases with
increasing IMAF’s.
19.
High boiling and melting points in a substance are caused by stronger forces of attraction.
20.
The vapor pressure of any substance is equal to 1 atm at its normal boiling point.
21.
Ionic solids are brittle with high melting points, and conduct electricity when molten or in solution.
22.
X = CH3-CH2-CH2-CH2-CH3
Y = CH3-CH2-CH2-CH2-OH
Z = HO-CH2-CH2-CH2-OH
Based on concepts of polarity and hydrogen bonding, which of the following sequences correctly lists
the compounds above in the order of their increasing solubility in water?
(A) Z < Y < X
(B) Y < Z < X
(C) Y < X < Z
(D) X < Z < Y
(E) X < Y < Z
23.
Explain why an egg takes longer to cook by boiling in Denver than in New York City.
The water boils at a lower temperature. There is a lower atmospheric pressure at higher
altitudes so the water requires less energy (lower temperature) to make its vapor pressure equal
to the lower atmospheric pressure to cause boiling.
(boiling – spontaneous phase change of liquid to gas when a liquid’s vapor pressure equals
atmospheric pressure)
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24.
When a solution is formed by adding some methanol, CH3OH, to water, processes that are
endothermic include which of the following?
I. Methanol molecules move water molecules apart as the methanol goes into solution.
II. Water molecules move methanol molecules apart as the methanol goes into solution.
III. Intermolecular attractions form between molecules of water and methanol as the methanol
goes into solution.
(A)
(B)
(C)
(D)
25.
solute-solute separation (endo)………………………CH3OH separates from CH3OH
I only
III only
I and II only
II and III only
solvent-solvent separation (endo)……………………H2O separates from H2O
solute-solvent intermolecular attraction (exo)……… CH3OH attract H2O
Why do endothermic processes occur if they don’t release heat (enthalpy, H)?
Endothermic processes absorb heat (enthalpy, H) which raises the energy of a system (not favored), but they may also
increase the disorder and randomness (entropy, S) of a system (favored) which does lower the overall energy of a system.
26.
Which one of the following substances would be the most soluble in CH3COOH (ethanoic acid)?
A. CBr4
B. CH2Cl2
C. C10H22
D. H2O
E. PH3
similar IMAFs (hydrogen bonding) between them makes them attract (mix) better
27.
Which of the following choices has the compounds correctly arranged in order of decreasing
solubility in water? (most soluble to least soluble)
A.
B.
C.
D.
E.
28.
CH4
CH3OH
LiF
CH3OH
CCl4
>
>
>
>
>
NH3
CCl4
NH3
CH4
CHCl3
> CHCl3
> CHCl3
> CHCl3
> LiF
> NH3
stronger IMAFs makes them attract
to polar water more (ion-dipole,
hydrogen bonding, dipole-dipole)
A sample of potassium chlorate (20.0 g) is dissolved in 99 g of water at 70 °C, with precautions taken
to avoid evaporation of any water. The solution is cooled to 20.0 °C and white crystalline precipitate is
observed. Describe and characterize the current solution as saturated, unsaturated, or supersaturated.
The solution is saturated (some solute could not dissolve). The solute is in a dynamic equilibrium (continues to dissolve
and precipitate out at equal rates) so that there is no net change in the amount of solute in solution.
4
29.
Define a supersaturated solution. Describe a simple test and the expected results to identify a solution
as supersaturated, saturated, or unsaturated.
A supersaturated solution contains more solute that it can typically hold at that temperature.
The seed crystal test: Add solute particles to the solution; if it’s supersaturated, all of the extra solute will crystallize out
of solution; if it’s saturated, the added solute will not dissolve; if it’s unsaturated, the added solute will dissolve.
30.
The concentration of potassium ion in a solution that contains 0.500 M potassium sulfate is 1.00 M.
H2O
K2SO4
0.500
0 M

2 K+ + SO42–
2(0.500)
0.500
1.00 M
0.500 M
31.
As the intermolecular attractions of liquids decrease, the vapor pressure of liquids increase causing
the boiling point of liquids to decrease.
32.
A closed rigid container contains water and N2(g) at equilibrium. Actions that would increase the
concentration on N2(g) in the water include which of the following?
I. Shaking the container vigorously
II. Raising the temperature of the water
III. Injecting more N2(g) into the container
(A) I only
33.
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
Which of the following has the lowest conductivity? Which has the highest conductivity?
(A) 0.1 M CuSO4
(B) 0.1 M KOH
(C) 0.1 M BaCl2
(D) 0.1 M HF
(E) 0.1 M HNO3
The best conductor would be BaCl2 b/c it would
dissociate into the most ions (3).
HF is a weak acid with less dissociation, fewer ions in
solution to conduct.
Element
Metallic Radius (pm)
Melting Point (oC)
Common Oxidation State
Au
144
1064
1+ , 3+
Cu
128
1085
1+ , 2+
Ag
144
961
1+
34.
To make Au stronger and harder, it is often alloyed with other metals, such as Cu and Ag.
Consider two alloys, one of Au and Cu and one of Au and Ag, each with the same mole fraction of Au.
If the Au/Cu alloy is harder than the Au/Ag alloy, then which of the following is the best explanation
based on the information in the table above?
(A) Cu has two common oxidation states, but Ag has only one.
(B) Cu has a higher melting point than Au has, but Ag has a lower melting point than Au has.
(C) Cu atoms are smaller than Ag atoms, thus they interfere more with the displacement of atoms in the alloy.
(D) Cu atoms are less polarizable than are Au or Ag atoms, thus Cu has weaker interparticle forces.
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[KEY] REVIEW Unit 5 Test Chp 11,13: Intermolecular Attractive Forces, Liquids, Solids, Solutions
Section II Free Response
Calculator Allowed
CLEARLY SHOW THE METHODS USED AND STEPS INVOLVED IN YOUR ANSWERS.
It is to your advantage to do this, because you may earn partial credit if you do and little or no credit if you do
not. Attention should be paid to significant figures.
1. Answer each of the following in terms of principles of molecular behavior and chemical concepts.
(a) The structures for benzene, C6H6, and nitroglycerin, C3H5N3O9, are shown below.
Identify the type(s) of intermolecular attractive forces in
(i)
pure benzene (1)
London dispersion forces
(ii)
pure nitroglycerin (1)
London dispersion forces and dipole-dipole interactions
(b) Nitroglycerin is soluble in water but benzene is not soluble in water. Explain. (2)
Benzene is nonpolar has only London dispersion forces and is therefore not soluble in a polar solvent like water.
Nitroglycerin has dipole-dipole forces and will attract water molecules thereby making it dissolve (or solvate) in
water.
(c) Indicate whether you agree or disagree with the statement in the box below. Support your answer with
a short explanation. (1)
When water boils, H2O molecules break apart to form
hydrogen molecules and oxygen molecules.
Disagree b/c boiling is a physical process (not chemical) involving only the separation of molecules by
overcoming the IMAF’s to break intermolecular bonds (not covalent intramolecular bonds) so the water
molecules remain intact.
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2. The condensed structural formulas of methylamine and ethane are shown below.
CH3NH2
CH3CH3
Methylamine
Ethane
(a) In the table, write the type(s) of intermolecular attractive force(s) that occur in each substance.
(2)
Substance
Boiling Point
Intermolecular Attractive Force(s)
Methylamine
267.2 K
London dispersion forces, dipole-dipole interactions,
hydrogen bonding
Ethane
184.5 K
London dispersion forces
(b) Use principles of intermolecular attractive forces to explain why ethane has a lower boiling point than
methylamine. (1)
Ethane has only dispersion forces but methylamine has dipole-dipole and hydrogen bonding. Therefore,
methylamine has stronger intermolecular attractive forces and requires more kinetic energy to overcome them
creating enough separation to become a gas. More kinetic energy required yields a higher boiling point than
that of ethane.
Answer the following questions related to a compound of sulfur, H2S .
(c) Two types of intermolecular forces present in liquid H2S are London dispersion forces and
dipole-dipole forces.
(i) Compare the strength of the London dispersion forces in liquid H2S to the strength of the
London dispersion forces in liquid H2O. Explain. (1)
The strength of the London forces in liquid H2S is greater than that of the London forces in liquid H2O.
The electron cloud of H2S has more electrons and is thus more polarizable than the electron cloud of the
H2O molecule.
(ii) Compare the strength of the dipole-dipole forces in liquid H2S to the strength of the
dipole-dipole forces in liquid H2O. Explain. (1)
The strength of the dipole-dipole forces in liquid H2S is weaker than that of the dipole-dipole forces in
liquid H2O. The net dipole moment of the H2S molecule is less than that of the H2O molecule.
This results from the lesser polarity of the H–S bond compared with that of the H–O bond (S is less
electronegative than O so there is a smaller difference of electronegativity between H–S than H–O).
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3. Account for each of the following observations about pairs of substances. In your answers, use
appropriate principles of chemical bonding and/or intermolecular forces. In each part, you must
include references to both substances.
(a) MgO melts at a much higher temperature (2,852oC) than NaF (993oC). (2)
MgO and NaF are both ionic solids. The +2 and 2 charges in MgO result in a greater attraction between ions
than the +1 and 1 charges in NaF. More energy is required to break the stronger ionic bonds.
(b) Si melts at a much higher temperature (1,410oC) than Cl2 (–101oC). (2)
Si is a network covalent solid with strong covalent bonds between atoms. Cl2 has weak London dispersion
forces between the molecules. Therefore, more energy is required to break the stronger bonds of Si than the
weak intermolecular forces of Cl2 .
(c) At 25C and 1.0 atm, ethane (C2H6) is a gas and hexane (C6H14) is a liquid. (2)
C2H6 and C6H14 both have London dispersion forces. The forces between C6H14 molecules are stronger
because the molecule has a larger electron cloud that is more polarizable.
(d) Structures of the dimethyl ether molecule and the ethanol molecule are shown below. The normal
boiling point of dimethyl ether is 250 K, whereas the normal boiling point of ethanol is 351 K.
Account for the difference in boiling points. (2)
The intermolecular forces of attraction among
molecules of dimethyl ether consist of London
(dispersion) forces and weak dipole-dipole
interactions. Ethanol has the same forces as
dimethyl ether, but can also form hydrogen bonds
between molecules
Hydrogen bonds are particularly strong intermolecular forces, so they require more energy to overcome during
the boiling process requiring a higher temperature to boil ethanol than is needed to boil dimethyl ether.
(e) Ammonia, NH3, is very soluble in water, whereas phosphine, PH3, is only moderately soluble in water. (2)
Ammonia has hydrogen bonding intermolecular attractive forces (IMAFs), whereas phosphine has dipoledipole interactions. Water also has hydrogen bonding IMAFs. Ammonia is more soluble in water than
phosphine because ammonia molecules can have more similar IMAFs with water (can form hydrogen bonds)
than phosphine. Phosphine does dissolve moderately well in water due to common dipole-dipole interactions,
but is not as strongly attracted to water as ammonia is by hydrogen bonding.
8
4. The structures of a water molecule and a crystal of LiCl(s) are represented above. A student prepares a
1.0 M solution by dissolving 4.2 g of LiCl(s) in enough water to make 100 mL of solution
(a) In the space provided below, show the interactions of the components of LiCl(aq) by making a
drawing that represents the different particles present in the solution. Base the particles in your
drawing on the particles shown in the representations above. Include only one formula unit of LiCl and
no more than ten molecules of water. Your drawing must include the following details. (2)
• identity of ions (symbol and charge)
• the arrangement and proper orientation of the particles in the solution
1 point is earned for a correctly drawn and
labeled particulate representation of the ions.
(must indicate that the smaller ion is Li+)
(separated into its ions and the ions are
correctly labeled with their respective identities
and charges)
H
H
H
O
Li+
O H
H
Cl
–
H
O H
H
O
1 point is earned for a correctly drawn
particulate representation of water molecules of
hydration surrounding the ions.
1 point is earned for correctly representing the
orientation of the water molecules of hydration
with the proper polarity.
The sketch should clearly show:
1. a clear representation of at least one Li+ ion and one Cl− ion separated from each other, labeled, and charged;
2. each ion surrounded by at least two H2O molecules (more solvent than solute); and
3. H2O molecules with the proper orientation around each ion (i.e., the O end of the water molecules closer to the
Li+ ion and the H end of the water molecules closer to the Cl− ion).
(b) The solvation of LiCl in water is an endothermic process. Describe the relative strengths of solutesolute, solvent-solvent, and solute-solvent attractions that are formed and broken in the solvation
process of LiCl in water.
The separations Li+ and Cl– and H2O---H2O are endothermic.
The formation of ion-dipole attractions of Li+ and Cl– with H2O are exothermic.
Since the overall process is endothermic, the solute-solute and solvent-solvent attractions must be stronger than
the solute-solvent attractions.
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Answer Key
1.
2006 #6
2.(a)
2.(b)
2.(c)
2005B #8(d)
2005B #8(e)
2009 #6(d)
3.(a)
3.(b)
3.(c)
3.(d)
3.(e)
2001
2001
2001
2008
2004
4.(a)
4.(b)
2013 #6(a)
(created)
#8(d)
#8(c)
#8(b)
#6(b)
#7(d)
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