CHAPTER TEN
LIQUIDS AND SOLIDS
For Review
1.
Intermolecular forces are the relatively weak forces between molecules that hold the
molecules together in the solid and liquid phases. Intramolecular forces are the forces within
a molecule. These are the covalent bonds in a molecule. Intramolecular forces (covalent
bonds) are much stronger than intermolecular forces.
Dipole forces are the forces that act between polar molecules. The electrostatic attraction
between the positive end of one polar molecule and the negative end of another is the dipole
force. Dipole forces are generally weaker than hydrogen bonding. Both of these forces are
due to dipole moments in molecules. Hydrogen bonding is given a separate name from dipole
forces because hydrogen bonding is a particularly strong dipole force. Any neutral molecule
that has a hydrogen covalently bonded to N, O, or F exhibits the relatively strong hydrogen
bonding intermolecular forces.
London dispersion forces are accidental-induced dipole forces. Like dipole forces, London
dispersion forces are electrostatic in nature. Dipole forces are the electrostatic forces between
molecules having a permanent dipole. London dispersion forces are the electrostatic forces
between molecules having an accidental or induced dipole. All covalent molecules (polar and
nonpolar) have London dispersion forces, but only polar molecules (those with permanent
dipoles) exhibit dipole forces.
As the size of a molecule increases, the strength of the London dispersion forces increases.
This is because, as the electron cloud about a molecule gets larger, it is easier for the electrons to be drawn away from the nucleus. The molecule is said to be more polarizable.
London dispersion (LD) < dipole-dipole < H bonding < metallic bonding, covalent network,
ionic.
Yes, there is considerable overlap. Consider some of the examples in Exercise 10.98.
Benzene (only LD forces) has a higher boiling point than acetone (dipole-dipole forces).
Also, there is even more overlap among the stronger forces (metallic, covalent, and ionic).
2.
a. Surface tension: the resistance of a liquid to an increase in its surface area.
b. Viscosity: the resistance of a liquid to flow.
c. Melting point: the temperature (at constant pressure) where a solid converts entirely to a
liquid as long as heat is applied. A more detailed definition is the temperature at which
the solid and liquid states have the same vapor pressure under conditions where the total
pressure is constant.
331
332
CHAPTER 10
LIQUIDS AND SOLIDS
d. Boiling point: the temperature (at constant pressure) where a liquid converts entirely to a
gas as long as heat is applied. The detailed definition is the temperature at which the
vapor pressure of the liquid is exactly equal to the external pressure.
e. Vapor pressure: the pressure of the vapor over a liquid at equilibrium.
As the strengths of intermolecular forces increase, surface tension, viscosity, melting point
and boiling point increase, while vapor pressure decreases.
3.
Solid: rigid; has fixed volume and shape; slightly compressible
Liquid: definite volume but no specific shape; assumes shape of the container; slightly
compressible
Gas: no fixed volume or shape; easily compressible
4.
a. Crystalline solid:
Regular, repeating structure
Amorphous solid: Irregular arrangement of atoms or molecules
b. Ionic solid:
Molecular solid:
c. Molecular solid:
Made up of discrete covalently bonded molecules held together in
the solid phase by weaker forces (LD, dipole, or hydrogen bonds).
Discrete, individual molecules
Network solid:
No discrete molecules; A network solid is one large molecule. The
forces holding the atoms together are the covalent bonds between
atoms.
d. Metallic solid:
Completely delocalized electrons, conductor of electricity (cations in
a sea of electrons)
Network solid:
5.
Made up of ions held together by ionic bonding
Localized electrons; Insulator or semiconductor
Lattice: a three-dimensional system of points designating the positions of the centers of the
components of a solid (atoms, ions, or molecules)
Unit cell: the smallest repeating unit of a lattice
A simple cubic unit cell has an atom, ion or molecule located at the eight corners of a cube.
There is one net atom per simple cubic unit cell. Because the atoms in the cubic unit cell are
assumed to touch along the cube edge, cube edge =  = 2r where r = radius of the atom. A
body-centered cubic unit cell has an atom, ion or molecule at the eight corners of a cube and
one atom, ion, or molecule located at the center of the cube. There are two net atoms per
body-centered cubic unit cell. Because the atoms in the cubic unit cell are assumed to touch
along the body diagonal of the cube, body diagonal = 3  = 4r where  = cube edge and r =
radius of atom. A face-centered cubic unit cell has an atom, ion, or molecule at the eight
CHAPTER 10
LIQUIDS AND SOLIDS
333
corners of a cube and an atom, ion, or molecule located at the six faces of the cube. There are
four net atoms per face-centered unit cell. Because the atoms in the cubic unit cell are
assumed to touch along the face diagonal of the cube, face diagonal = 2  = 4r.
6.
Closest packing: the packing of atoms (uniform, hard spheres) in a manner that most
efficiently uses the available space with the least amount of empty space. The two types of
closest packing are hexagonal closest packing and cubic closest packing. In both closest
packed arrangements, the atoms (spheres) are packed in layers. The difference between the
two closest packed arrangements is the ordering of the layers. Hexagonal closest packing has
the third layer directly over the first layer forming a repeating layer pattern of abab… In
cubic closest packing the layer pattern is abcabc… The unit cell for hexagonal closest
packing is a hexagonal prism. See Fig. 10.14 for an illustration of the hexagonal prism unit
cell. The unit cell for cubic closest packing is the face-centered cubic unit cell.
7.
Conductor:
The energy difference between the filled and unfilled molecular orbitals is
minimal. We call this energy difference the band gap. Because the band gap
is minimal, electrons can easily move into the conduction bands (the unfilled
molecular orbitals).
Insulator:
Large band gap; Electrons do not move from the filled molecular orbitals to
the conduction bands since the energy difference is large.
Semiconductor: Small band gap; The energy difference between the filled and unfilled molecular orbitals is smaller than in insulators, so some electrons can jump into
the conduction bands. The band gap, however, is not as small as with conductors, so semiconductors have intermediate conductivity.
a. As the temperature is increased, more electrons in the filled molecular orbitals have
sufficient kinetic energy to jump into the conduction bands (the unfilled molecular
orbitals).
b. A photon of light is absorbed by an electron which then has sufficient energy to jump into
the conduction bands.
c. An impurity either adds electrons at an energy near that of the conduction bands (n-type)
or creates holes (unfilled energy levels) at energies in the previously filled molecular
orbitals (p-type). Both n-type and p-type semiconductors increase conductivity by
creating an easier path for electrons to jump from filled to unfilled energy levels.
In conductors, electrical conductivity is inversely proportional to temperature. Increases in
temperature increase the motions of the atoms, which gives rise to increased resistance
(decreased conductivity). In a semiconductor, electrical conductivity is directly proportional
to temperature. An increase in temperature provides more electrons with enough kinetic
energy to jump from the filled molecular orbitals to the conduction bands, increasing
conductivity.
To produce an n-type semiconductor, dope Ge with a substance that has more than 4 valence
electrons, e.g., a group 5A element. Phosphorus or arsenic are two substances which will
produce n-type semiconductors when they are doped into germanium. To produce a p-type
semiconductor, dope Ge with a substance that has fewer than 4 valence electrons, e.g., a
334
CHAPTER 10
LIQUIDS AND SOLIDS
group 3A element. Gallium or indium are two substances which will produce p-type semiconductors when they are doped into germanium.
8.
The structures of most binary ionic solids can be explained by the closest packing of spheres.
Typically, the larger ions, usually the anions, are packed in one of the closest packing
arrangements, and the smaller cations fit into holes among the closest packed anions. There
are different types of holes within the closest packed anions which are determined by the
number of spheres that form them. Which of the three types of holes are filled usually
depends on the relative size of the cation to the anion. Ionic solids will always try to
maximize electrostatic attractions among oppositely charged ions and minimize the
repulsions among ions with like charges.
The structure of sodium chloride can be described in terms of a cubic closest packed array of
Cl ions with Na+ ions in all of the octahedral holes. An octahedral hole is formed between 6
Cl anions. The number of octrahedral holes is the same as the number of packed ions. So in
the face-centered unit cell of sodium chloride, there are 4 net Cl ions and 4 net octahedral
holes. Because the stoichiometry dictates a 1:1 ratio between the number of Cl anions and
Na+ cations, all of the octahedral holes must be filled with Na+ ions.
In zinc sulfide, the sulfide anions also occupy the lattice points of a cubic closest packing
arrangement. But instead of having the cations in octahedral holes, the Zn2+ cations occupy
tetrahedral holes. A tetrahedral hole is the empty space created when four spheres are packed
together. There are twice as many tetrahedral holes as packed anions in the closest packed
structure. Therefore, each face-centered unit cell of sulfide anions contains 4 net S2 ions and
8 net tetrahedral holes. For the 1:1 stoichiometry to work out, only one-half of the tetrahedral
holes are filled with Zn2+ ions. This gives 4 S2 ions and 4 Zn2+ ions per unit cell for an
empirical formula of ZnS.
9.
a. Evaporation: process where liquid molecules escape the liquid’s surface to form a gas.
b. Condensation: process where gas molecules hit the surface of a liquid and convert to a
liquid.
c. Sublimation: process where a solid converts directly to a gas without passing through the
liquid state.
d. Boiling: the temperature and pressure at which a liquid completely converts to a gas as
long as heat is applied.
e. Melting: temperature and pressure at which a solid completely converts to a liquid as
long as heat is applied.
f.
Enthalpy of vaporization (Hvap): the enthalpy change that occurs at the boiling point
when a liquid converts into a gas.
g. Enthalpy of fusion (Hfus): the enthalpy change that occurs at the melting point when a
solid converts into a liquid.
CHAPTER 10
LIQUIDS AND SOLIDS
335
h. Heating curve: a plot of temperature versus time as heat is applied at a constant rate to
some substance.
Fusion refers to a solid converting to a liquid, and vaporization refers to a liquid converting to
a gas. Only a fraction of the hydrogen bonds in ice are broken in going from the solid phase
to the liquid phase. Most of the hydrogen bonds in water are still present in the liquid phase
and must be broken during the liquid to gas phase transition. Thus, the enthalpy of
vaporization is much larger than the enthalpy of fusion because more intermolecular forces
are broken during the vaporization process.
A volatile liquid is one that evaporates relatively easily. Volatile liquids have large vapor
pressures because the intermolecular forces that prevent evaporation are relatively weak.
10.
See Fig. 10.49 and 10.52 for the phase diagrams of H2O and CO2. Most substances exhibit
only three different phases: solid, liquid, and gas. This is true for H2O and CO2. Also typical
of phase diagrams is the positive slopes for both the liquid-gas equilibrium line and the solidgas equilibrium line. This is also true for both H2O and CO2. The solid-liquid equilibrium line
also generally has a positive slope. This is true for CO2, but not for H2O. In the H2O phase
diagram, the slope of the solid-liquid line is negative. The determining factor for the slope of
the solid-liquid line is the relative densities of the solid and liquid phases. The solid phase is
denser than the liquid phase in most substances; for these substances, the slope of the solidliquid equilibrium line is positive. For water, the liquid phase is denser than the solid phase
which corresponds to a negative sloping solid-liquid equilibrium line. Another difference
between H2O and CO2 is the normal melting points and normal boiling points. The term
normal just dictates a pressure of 1 atm. H2O has a normal melting point (0C) and a normal
boiling point (100C), but CO2 does not. At 1 atm pressure, CO2 only sublimes (goes from
the solid phase directly to the gas phase). There are no temperatures at 1 atm for CO 2 where
the solid and liquid phases are in equilibrium or where the liquid and gas phases are in
equilibrium. There are other differences, but those discussed above are the major ones.
The relationship between melting points and pressure is determined by the slope of the solidliquid equilibrium line. For most substances (CO2 included), the positive slope of the solidliquid line shows a direct relationship between the melting point and pressure. As pressure
increases, the melting point increases. Water is just the opposite since the slope of the solidliquid line in water is negative. Here the melting point of water is inversely related to the
pressure.
For boiling points, the positive slope of the liquid-gas equilibrium line indicates a direct
relationship between the boiling point and pressure. This direct relationship is true for all
substances including H2O and CO2.
The critical temperature for a substance is defined as the temperature above which the vapor
cannot be liquefied no matter what pressure is applied. The critical temperature, like the
boiling point temperature, is directly related to the strength of the intermolecular forces. Since
H2O exhibits relatively strong hydrogen bonding interactions and CO2 only exhibits London
dispersion forces, one would expect a higher critical temperature for H2O than for CO2.
336
CHAPTER 10
LIQUIDS AND SOLIDS
Questions
12.
C25H52 has the stronger intermolecular forces because it has the higher boiling point. Even
though C25H52 is nonpolar, it is so large that its London dispersion forces are much stronger
than the sum of the London dispersion and hydrogen bonding interactions found in H2O.
13.
Atoms have an approximately spherical shape (on the average). It is impossible to pack
spheres together without some empty space among the spheres.
14.
Critical temperature:
Critical pressure:
The temperature above which a liquid cannot exist, i.e., the gas
cannot be liquified by increased pressure.
The pressure that must be applied to a substance at its critical
temperature to produce a liquid.
The kinetic energy distribution changes as one raises the temperature (T4 > Tc > T3 > T2 >
T1). At the critical temperature, Tc, all molecules have kinetic energies greater than the
intermolecular forces, F, and a liquid can't form. Note: The distributions above are not to
scale.
15.
Evaporation takes place when some molecules at the surface of a liquid have enough energy
to break the intermolecular forces holding them in the liquid phase. When a liquid
evaporates, the molecules that escape have high kinetic energies. The average kinetic energy
of the remaining molecules is lower, thus, the temperature of the liquid is lower.
16.
A crystalline solid will have the simpler diffraction pattern because a regular, repeating
arrangement is necessary to produce planes of atoms that will diffract the X-rays in regular
patterns. An amorphous solid does not have a regular repeating arrangement and will
produce a complicated diffraction pattern.
17.
An alloy is a substance that contains a mixture of elements and has metallic properties. In a
substitutional alloy, some of the host metal atoms are replaced by other metal atoms of
similar size, e.g., brass, pewter, plumber’s solder. An interstitial alloy is formed when some
of the interstices (holes) in the closest packed metal structure are occupied by smaller atoms,
e.g., carbon steels.
18.
Equilibrium:
There is no change in composition; the vapor pressure is constant.
Dynamic:
Two processes, vapor → liquid and liquid → vapor, are both occurring but
with equal rates so the composition of the vapor is constant.
CHAPTER 10
19.
LIQUIDS AND SOLIDS
337
a. As the strength of the intermolecular forces increase, the rate of evaporation decreases.
b. As temperature increases, the rate of evaporation increases.
c. As surface area increases, the rate of evaporation increases.
20.
C2H5OH(l) → C2H5OH(g) is an endothermic process. Heat is absorbed when liquid ethanol
vaporizes; the internal heat from the body provides this heat which results in the cooling of
the body.
21.
Sublimation will occur allowing water to escape as H2O(g).
22.
The phase change, H2O(g) → H2O(l), releases heat that can cause additional damage. Also
steam can be at a temperature greater than 100°C.
23.
The strength of intermolecular forces determines relative boiling points. The types of
intermolecular forces for covalent compounds are London dispersion forces, dipole forces,
and hydrogen bonding. Because the three compounds are assumed to have similar molar mass
and shape, the strength of the London dispersion forces will be about equal between the three
compounds. One of the compounds will be nonpolar so it only has London dispersion forces.
The other two compounds will be polar so they have additional dipole forces and will boil at
a higher temperature than the nonpolar compound. One of the polar compounds will have an
H covalently bonded to either N, O, or F. This gives rise to the strongest type of covalent
intermolecular forces, hydrogen bonding. The compound which hydrogen bonds will have the
highest boiling point while the polar compound with no hydrogen bonding will boil at a
temperature in the middle of the other compounds.
24.
a. Both forms of carbon are network solids. In diamond, each carbon atom is surrounded by
a tetrahedral arrangement of other carbon atoms to form a huge molecule. Each carbon
atom is covalently bonded to four other carbon atoms.
The structure of graphite is based on layers of carbon atoms arranged in fused sixmembered rings. Each carbon atom in a particular layer of graphite is surrounded by
three other carbons in a trigonal planar arrangement. This requires sp 2 hybridization.
Each carbon has an unhybridized p atomic orbital; all of these p orbitals in each sixmembered ring overlap with each other to form a delocalized  electron system.
b. Silica is a network solid having an empirical formula of SiO2. The silicon atoms are
singly bonded to four oxygens. Each silicon atom is at the center of a tetrahedral
arrangement of oxygen atoms which are shared with other silicon atoms. The structure of
silica is based on a network of SiO4 tetrahedra with shared oxygen atoms, rather than
discrete SiO2 molecules.
Silicates closely resemble silica. The structure is based on interconnected SiO4 tetrahedra.
However, in contrast to silica, where the O/Si ratio is 2:1, silicates have O/Si ratios
greater than 2:1 and contain silicon-oxygen anions. To form a neutral solid silicate, metal
cations are needed to balance the charge. In other words, silicates are salts containing
metal cations and polyatomic silicon-oxygen anions.
338
CHAPTER 10
LIQUIDS AND SOLIDS
When silica is heated above its melting point and cooled rapidly, an amorphous
(disordered) solid called glass results. Glass more closely resembles a very viscous
solution than it does a crystalline solid. To affect the properties of glass, several different
additives are thrown into the mixture. Some of these additives are Na2CO3, B2O3, and
K2O, with each compound serving a specific purpose relating to the properties of the
glass.
25.
a. Both CO2 and H2O are molecular solids. Both have an ordered array of the individual
molecules, with the molecular units occupying the lattice points. A difference within
each solid lattice is the strength of the intermolecular forces. CO2 is nonpolar and only
exhibits London dispersion forces. H2O exhibits the relatively strong hydrogen bonding
interactions. The differences in strength is evidenced by the solid phase changes that
occur at 1 atm. CO2(s) sublimes at a relatively low temperature of 78C. In sublimation,
all of the intermolecular forces are broken. However, H2O(s) doesn’t have a phase change
until 0C, and in this phase change from ice to water, only a fraction of the
intermolecular forces are broken. The higher temperature and the fact that only a portion
of the intermolecular forces are broken are attributed to the strength of the intermolecular
forces in H2O(s) as compared to CO2(s).
Related to the intermolecular forces are the relative densities of the solid and liquid
phases for these two compounds. CO2(s) is denser than CO2(l) while H2O(s) is less dense
than H2O(l). For CO2(s), the molecules pack together as close as possible, hence solids
are usually more dense than the liquid phase. For H2O, each molecule has two lone pairs
and two bonded hydrogen atoms. Because of the equal number of lone pairs and OH
bonds, each H2O molecule can form two hydrogen bonding interactions to other H2O
molecules. To keep this symmetric arrangement (which maximizes the hydrogen bonding
interactions), the H2O(s) molecules occupy positions that create empty space in the
lattice. This translates into a smaller density for H2O(s) as compared to H2O(l).
b. Both NaCl and CsCl are ionic compounds with the anions at the lattice points of the unit
cells and the cations occupying the empty spaces created by anions (called holes). In
NaCl, the Cl anions occupy the lattice points of a face-centered unit cell with the Na+
cations occupying the octahedral holes. Octahedral holes are the empty spaces created by
six Cl ions. CsCl has the Cl ions at the lattice points of a simple cubic unit cell with the
Cs+ cations occupying the middle of the cube.
26.
Because silicon carbide is made from Group 4A elements and because it is extremely hard,
one would expect SiC to form a covalent network structure similar to diamond.
27.
The mathematical equation that relates the vapor pressure of a substance to temperature is:
ln Pvap = 
y
ΔH vap  1 
  +C
R T
m
x +b
As shown above, this equation is in the form of the straight line equation. If one plots ln P vap
vs. 1/T, the slope of the straight line is Hvap/R. Because Hvap is always positive, the slope
of the straight line will be negative.
CHAPTER 10
28.
LIQUIDS AND SOLIDS
339
The typical phase diagram for a substance shows three phases and has a positive sloping
solid-liquid equilibrium line (water is atypical). A sketch of the phase diagram for I2 would
look like this:
s
l
g
P 90
torr
115oC
T
Statements a and e are true. For statement a, the liquid phase is always more dense than the
gaseous phase (gases are mostly empty space). For statement e, because the triple point is at
90 torr, the liquid phase cannot exist at any pressure less than 90 torr, no matter what the
temperature. For statements b, c, and d, examine the phase diagram to prove to yourself that
they are false.
Exercises
Intermolecular Forces and Physical Properties
29.
30.
Ionic compounds have ionic forces. Covalent compounds all have London Dispersion (LD)
forces, while polar covalent compounds have dipole forces and/or hydrogen bonding forces.
For H bonding forces, the covalent compound must have either a NH, OH or FH bond in
the molecule.
a. LD only
b. dipole, LD
d. ionic
e. LD only (CH4 in a nonpolar covalent compound.)
f.
g. ionic
dipole, LD
c. H bonding, LD
a. ionic
b. LD mostly; C‒F bonds are polar, but polymers like teflon are so large the LD forces are
the predominant intermolecular forces.
31.
c. LD
d. dipole, LD
f.
g. LD
dipole, LD
e. H bonding, LD
a. OCS; OCS is polar and has dipole-dipole forces in addition to London dispersion (LD)
forces. All polar molecules have dipole forces. CO2 is nonpolar and only has LD forces.
To predict polarity, draw the Lewis structure and deduce whether the individual bond
dipoles cancel.
340
CHAPTER 10
LIQUIDS AND SOLIDS
b. SeO2; Both SeO2 and SO2 are polar compounds, so they both have dipole forces as well
as LD forces. However, SeO2 is a larger molecule, so it would have stronger LD forces.
c. H2NCH2CH2NH2; More extensive hydrogen bonding is possible.
d. H2CO; H2CO is polar while CH3CH3 is nonpolar. H2CO has dipole forces in addition to
LD forces.
e. CH3OH; CH3OH can form relatively strong H bonding interactions, unlike H2CO.
32.
Ar exists as individual atoms which are held together in the condensed phases by London
dispersion forces. The molecule which will have a boiling point closest to Ar will be a
nonpolar substance with about the same molar mass as Ar (39.95 g/mol); this same size
nonpolar substance will have about equivalent strength of London dispersion forces. Of the
choices, only Cl2 (70.90 g/mol) and F2 (38.00 g/mol) are nonpolar. Because F2 has a molar
mass closest to that of Ar, one would expect the boiling point of F2 to be close to that of Ar.
33.
a. Neopentane is more compact than n-pentane. There is less surface area contact among
neopentane molecules. This leads to weaker LD forces and a lower boiling point.
b. HF is capable of H bonding; HCl is not.
c. LiCl is ionic, and HCl is a molecular solid with only dipole forces and LD forces. Ionic
forces are much stronger than the forces for molecular solids.
d. n-Hexane is a larger molecule, so it has stronger LD forces.
34.
Ethanol, C2H6O, has 2(4) + 6(1) + 6 = 20 valence electrons.
H
H
H
C
C
H
H
O
H
Exhibits H-bonding and London
dispersion forces.
Dimethyl ether, C2H6O, also has 20 valence electrons. It has a Lewis structure of:
H
H
O
H
C
C
H
H
H
Exhibits dipole and London
dispersion forces, but no hydrogen
bonding since it has no H covalently
bonded to the O.
CHAPTER 10
LIQUIDS AND SOLIDS
341
Propane, C3H6, has 3(4) + 6(1) = 18 valence electrons.
H
H
H
H
C
C
C
H
H
H
H
Propane only has relatively nonpolar bonds
so it is nonpolar. Propane exhibits only
London dispersion forces.
The three compounds have similar molar mass so the strength of the London dispersion
forces will be approximately equivalent. Because dimethyl ether has additional dipole forces,
it will boil at a higher temperature than propane. The compound with the highest boiling
point is ethanol since it exhibits relatively strong hydrogen bonding forces. The correct
matching of boiling points is:
ethanol, 78.5C; dimethyl ether, 23C; propane, 42.1C
35.
Boiling points and freezing points are assumed directly related to the strength of the
intermolecular forces, while vapor pressure is inversely related to the strength of the
intermolecular forces.
a. HBr; HBr is polar, while Kr and Cl2 are nonpolar. HBr has dipole forces unlike Kr and
Cl2.
b. NaCl; Ionic forces are much stronger than molecular forces.
c. I2; All are nonpolar, so the largest molecule (I2) will have the strongest LD forces and the
lowest vapor pressure.
d. N2; Nonpolar and smallest, so has the weakest intermolecular forces.
e. CH4; Smallest, nonpolar molecule so has the weakest LD forces.
f.
HF; HF can form relatively strong H bonding interactions unlike the others.
g. CH3CH2CH2OH; H bonding, unlike the others, so has strongest intermolecular forces.
36.
a. CBr4; Largest of these nonpolar molecules so has strongest LD forces.
b. F2; Ionic forces in LiF are much stronger than the covalent forces in F2 and HCl. HCl has
dipole forces that the nonpolar F2 does not exhibit; so F2 has the weakest intermolecular
forces and the lowest freezing point.
c. CH3CH2OH; Can form H bonding interactions unlike the others.
d. H2O2; HO-OH structure produces stronger H bonding interactions than HF, so has
greatest viscosity.
e.
H2CO; H2CO is polar so has dipole forces, unlike the other nonpolar covalent
compounds.
342
CHAPTER 10
f.
LIQUIDS AND SOLIDS
I2; I2 has only LD forces while CsBr and CaO have much stronger ionic forces. I2 has
weakest intermolecular forces so has smallest ΔHfusion.
Properties of Liquids
37.
The attraction of H2O for glass is stronger than the H2O‒H2O attraction. The miniscus is
concave to increase the area of contact between glass and H2O. The Hg‒Hg attraction is
greater than the Hg‒glass attraction. The miniscus is convex to minimize the Hg‒glass
contact.
38.
A molecule at the surface of a waterdrop is subject to attractions only by molecules below it
and to each side. The effect of this uneven pull on the surface molecules tends to draw them
into the body of the liquid and causes the droplet to assume the shape that has the minimum
surface area, a sphere.
39.
The structure of H2O2 is H‒O‒O‒H, which produces greater hydrogen bonding than water.
Long chains of hydrogen‒bonded H2O2 molecules then get tangled together.
40.
CO2 is a gas at room temperature. As mp and bp increase, the strength of the intermolecular
forces also increases. Therefore, the strength of forces is CO2 < CS2 < CSe2. From a structural
standpoint this is expected. All three are linear, nonpolar molecules. Thus, only London
dispersion forces are present. Since the molecules increase in size from CO2 < CS2 < CSe2,
the strength of the intermolecular forces will increase in the same order.
Structures and Properties of Solids
41.
nλ = 2d sin θ, d =
42.
d=
43.
=
44.
nλ = 2d sin θ, d =
nλ
2  154 pm
=
= 408 pm = 4.08 × 10 10 m
2 sin θ
2  sin 22.20
2  1.36  10 10 m  sin 15.0 o
2 d sin θ
=
= 7.04 × 10 11 m = 0.704 
1
n
sin θ =
45.
nλ
1  154 pm
=
= 313 pm = 3.13 × 10 10 m
2 sin θ
2  sin 14.22
1  2.63 
nλ
=
= 4.91  = 4.91 × 10 10 m = 491 pm
2  sin 15.55 
2 sin θ
2  2.63 
nλ
=
= 0.536, θ = 32.4°
2  4.91 
2d
A cubic closest packed structure has a face-centered cubic unit cell. In a face-centered cubic
unit, there are:
8 corners ×
1/ 8 atom
1 / 2 atom
= 4 atoms
 6 faces
corner
face
CHAPTER 10
LIQUIDS AND SOLIDS
343
The atoms in a face-centered cubic unit cell touch along the face diagonal of the cubic unit
cell. Using the Pythagorean formula where l = length of the face diagonal and r = radius of
the atom:
l2 + l2 = (4r)2
4r
l
2 l2 = 16 r2
l=r 8
l
l=r
8 = 197 × 10 12 m ×
8 = 5.57 × 10 10 m = 5.57 × 10 8 cm
Volume of a unit cell = l 3 = (5.57 × 10 8 cm)3 = 1.73 × 10 22 cm3
Mass of a unit cell = 4 Ca atoms ×
density =
46.
1 mol Ca
40.08 g Ca
= 2.662 × 10 22 g Ca

23
mol Ca
6.022  10 atoms
mass
2.662  10 22 g

= 1.54 g/cm3
 22
3
volume 1.73  10 cm
There are 4 Ni atoms in each unit cell: For a unit cell:
density =
mass
= 6.84 g/cm3 =
volume
4 Ni atoms 
1 mol Ni
58.69 g Ni

23
mol Ni
6.022  10 atoms
3
l
Solving: l = 3.85 × 10 8 cm = cube edge length
For a face centered cube:
(4r)2 = l 2 + l 2 = 2 l 2
r 8 = l, r = l / 8
4r
l
r = 3.85 × 10 8 cm / 8
r = 1.36 × 10 8 cm = 136 pm
l
344
47.
CHAPTER 10
LIQUIDS AND SOLIDS
The unit cell for cubic closest packing is the face-centered unit cell. The volume of a unit cell
is:
V = l 3 = (492 × 10 10 cm)3 = 1.19 × 10 22 cm3
There are 4 Pb atoms in the unit cell, as is the case for all face-centered cubic unit cells. The
mass of atoms in a unit cell is:
mass = 4 Pb atoms ×
density =
1 mol Pb
207.2 g Pb
= 1.38 × 10 21 g

23
mol Pb
6.022  10 atoms
mass
1.38  10 21 g

= 11.6 g/cm3
volume 1.19  10  22 cm3
From Exercise 45, the relationship between the cube edge length, l, and the radius of an atom
in a face-centered unit cell is: l = r 8 .
r=
l
=
8
48.
492 pm
8
= 174 pm = 1.74 × 10 10 m
A face-centered cubic unit cell contains 4 atoms. For a unit cell:
mass of X = volume × density = (4.09 × 10 8 cm)3 × 10.5 g/cm3 = 7.18 × 10 22 g
mol X = 4 atoms X ×
Molar mass =
49.
1 mol X
= 6.642 × 10 24 mol X
6.022  10 23 atoms
7.18  10 22 g X
= 108 g/mol; The metal is silver (Ag).
6.642  10  24 mol X
For a body-centered unit cell: 8 corners ×
1 / 8 Ti
+ Ti at body center = 2 Ti atoms
corner
All body-centered unit cells have 2 atoms per unit cell. For a unit cell:
2 atoms Ti 
density = 4.50 g/cm3 =
1 mol Ti
47.88 g Ti

mol Ti
6.022  10 23 atoms
3
l
Solving: l = edge length of unit cell = 3.28 × 10 8 cm = 328 pm
Assume Ti atoms just touch along the body diagonal of the cube, so body diagonal =
4 × radius of atoms = 4r.
CHAPTER 10
LIQUIDS AND SOLIDS
345
The triangle we need to solve is:
4r
l
l 2
3.28 x 10 -8 cm
(3.28 x 10 -8cm)  2
(4r)2 = (3.28 × 10 8 cm)2 + [(3.28 × 10 8 cm)
2 ]2, r = 1.42 × 10 8 cm = 142 pm
For a body-centered unit cell (bcc), the radius of the atom is related to the cube edge length
by: 4r = l 3 or l = 4r/ 3 .
50.
From Exercise 10.49:
4r
16 r2 = l 2 + 2 l 2
l = 4r/ 3 = 2.309 r
l
l = 2.309 (222 pm) = 513 pm = 5.13 × 10 8 cm
l 2
In a bcc, there are 2 atoms/unit cell. For a unit cell:
density =
51.
mass

volume
2 atoms Ba 
1 mol Ba
137.3 g Ba

23
mol Ba
3.38 g
6.022  10 atoms

8
3
(5.13  10 cm)
cm3
If a face-centered cubic structure, then 4 atoms/unit cell and from Exercise 10.45:
2 l 2 = 16 r2
4r
l
l = r 8 = (144 pm)
8 = 407 pm
l = 4.07 × 10 10 m = 4.07 × 10 8 cm
l
4 atoms Au 
density =
1 mol Au
197 .0 g Au

23
mol Au
6.022  10 atoms
= 19.4 g/cm3
8
3
(4.07  10 cm)
346
CHAPTER 10
LIQUIDS AND SOLIDS
If a body-centered cubic structure, then 2 atoms/unit cell and from Exercise 10.49:
16 r2 = l 2 + 2 l 2
4r
l
l = 4r/ 3 = 333 pm = 3.33 × 10 10 m
l = 3.33 × 10 8 cm
l 2
2 atoms Au 
density =
1 mol Au
197 .0 g Au

23
mol Au
6.022  10 atoms
= 17.7 g/cm3
8
3
(3.33  10 cm)
The measured density is consistent with a face-centered cubic unit cell.
52.
If face-centered cubic:
l = r 8 = (137 pm)
8 = 387 pm = 3.87 × 10 8 cm
4 atoms W 
density =
1 mol
183 .9 g W

23
mol
6.022  10 atoms
= 21.1 g/cm3
8
3
(3.87  10 cm)
If body-centered cubic:
l =
4r
3
=
4  137 pm
3
2 atoms W 
density =
= 316 pm = 3.16 × 10 8 cm
1 mol
183 .9 g W

mol
6.022  10 23 atoms
= 19.4 g/cm3
8
3
(3.16  10 cm)
The measured density is consistent with a body-centered unit cell.
53.
In a face-centered unit cell (ccp structure), the atoms touch along the face diagonal:
(4r)2 = l 2 + l2
4r
l
l=r 8
Vcube = l 3 = ( r 8 )3 = 22.63 r3
There are four atoms in a face-centered cubic cell (see Exercise 10.45). Each atom has a
volume of 4/3 πr3.
CHAPTER 10
LIQUIDS AND SOLIDS
Vatoms = 4 ×
4
3
347
πr3 = 16.76 r3
Vatoms 16 .76 r 3

= 0.7406 or 74.06% of the volume of each unit cell is occupied by
Vcube 22 .63 r 3
atoms.
So,
In a simple cubic unit cell, the atoms touch along the cube edge (l):
2(radius) = 2r = l
2r
l
Vcube = l3 = (2r)3 = 8 r3
There is one atom per simple cubic cell (8 corner atoms × 1/8 atom per corner = 1 atom/unit
cell). Each atom has an assumed volume of 4/3 πr3 = volume of a sphere.
Vatom =
4
3
πr3 = 4.189 r3
Vatom 4.189 r 3

= = 0.5236 or 52.36% of the volume of each unit cell is occupied by
Vcube
8 r3
atoms.
So,
A cubic closest packed structure packs the atoms much more efficiently than a simple cubic
structure.
54.
From Exercise 10.49, a body-centered unit cell contains 2 net atoms, and the length of a cube
edge (l) is related to the radius of the atom (r) by the equation l = 4r/ 3 .
Volume of unit cell = l3 = (4 r/ 3 )3 = 12.32 r3
Volume of atoms in unit cell = 2 ×
So:
4
3
π r3 = 8.378 r3
Vatoms 8.378 r 3

= 0.6800 = 68.00% occupied
Vcube 12 .32 r 3
To determine the radius of the Fe atoms, we need to determine the cube edge length (l).

1 mol Fe
55.85 g Fe  1 cm3


Volume of unit cell =  2 Fe atoms 
mol Fe  7.86 g
6.022  10 23 atoms

= 2.36 × 10 23 cm3
Volume = l 3 = 2.36 × 10 23 cm3, l = 2.87 × 10 8 cm
l = 4r/ 3 , r = l 3 / 4 = 2.87 × 10 8 cm ×
3 / 4 = 1.24 × 10 8 cm
348
CHAPTER 10
LIQUIDS AND SOLIDS
55.
Doping silicon with phosphorus produces an n-type semiconductor. The phosphorus adds
electrons at energies near the conduction band of silicon. Electrons do not need as much
energy to move from filled to unfilled energy levels so conduction increases. Doping silicon
with gallium produces a p-type semiconductor. Because gallium has fewer valence electrons
than silicon, holes (unfilled energy levels) at energies in the previously filled molecular
orbitals are created, which induces greater electron movement (greater conductivity).
56.
A rectifier is a device that produces a current that flows in one direction from an alternating
current which flows in both directions. In a p-n junction, a p-type and an n-type semiconductor are connected. The natural flow of electrons in a p-n junction is for the excess
electrons in the n-type semiconductor to move to the empty energy levels (holes) of the ptype semiconductor. Only when an external electric potential is connected so that electrons
flow in this natural direction will the current flow easily (forward bias). If the external
electric potential is connected in reverse of the natural flow of electrons, no current flows
through the system (reverse bias). A p-n junction only transmits a current under forward bias,
thus converting the alternating current to direct current.
57.
In has fewer valence electrons than Se, thus, Se doped with In would be a p-type
semiconductor.
58.
To make a p-type semiconductor we need to dope the material with atoms that have fewer
valence electrons. The average number of valence electrons is four when 50-50 mixtures of
group 3A and group 5A elements are considered. We could dope with more of the Group 3A
element or with atoms of Zn or Cd. Cadmium is the most common impurity used to produce
p-type GaAs semiconductors. To make an n-type GaAs semiconductor, dope with an excess
group 5A element or dope with a Group 6A element such as sulfur.
59.
Egap = 2.5 eV × 1.6 × 10 19 J/eV = 4.0 × 10 19 J; We want Egap = Elight, so:
λ 
hc (6.63  10 34 J s) (3.00  10 8 m / s)
=
= 5.0 × 10 7 m = 5.0 × 102 nm
4.0  10 19 J
E
hc (6.63  10 34 J s) (2.998  10 8 m / s)

= 2.72 × 10 19 J = energy of band gap
9
λ
730 .  10 m
60.
E=
61.
a. 8 corners ×
1/ 8 Cl
1 / 2 Cl
+ 6 faces ×
= 4 Cl ions
face
corner
1/ 4 Na
+ 1 Na at body center = 4 Na ions; NaCl is the formula.
edge
1/ 8 Cl
b. 1 Cs ion at body center; 8 corners ×
= 1 Cl ion; CsCl is the formula.
corner
12 edges ×
c. There are 4 Zn ions inside the cube.
8 corners ×
1/ 8 S
1/ 2 S
+ 6 faces ×
= 4 S ions; ZnS is the formula.
face
corner
CHAPTER 10
LIQUIDS AND SOLIDS
d. 8 corners ×
4 faces ×
62.
349
1/ 8 Ti
+ 1 Ti at body center = 2 Ti ions
corner
1/ 2 O
+ 2 O inside cube = 4 O ions; TiO2 is the formula.
face
Both As ions are inside the unit cell. 8 corners ×
1/ 8 Ni
1 / 4 Ni
+ 4 edges ×
= 2 Ni ions
corner
edge
The unit cell contains 2 ions of Ni and 2 ions of As which gives a formula of NiAs.
63.
There is one octahedral hole per closest packed anion in a closest packed structure. If half of
the octahedral holes are filled, there is a 2:1 ratio of fluoride ions to cobalt ions in the crystal.
The formula is CoF2.
64.
There are 2 tetrahedral holes per closest packed anion. Let f = fraction of tetrahedral holes
filled by the cations.
Na2O: cation to anion ratio =
2
2f
=
, f = 1; All of the tetrahedral holes are filled by Na+
1
1
cations.
CdS: cation to anion ratio =
1
2f
1
=
, f = ; 1/2 of the tetrahedral holes are filled by
1
1
2
Cd2+ cations.
ZrI4: cation to anion ratio =
1
2f
1
=
, f = ; 1/8 of the tetrahedral holes are filled by
4
1
8
Zr4+ cations.
65.
In a cubic closest packed array of anions, there are twice the number of tetrahedral holes as
anions present and an equal number of octahedral holes as anions present. A cubic closest
packed array of sulfide ions will have 4 S2 ions, 8 tetrahedral holes, and 4 octahedral holes.
In this structure we have 1/8(8) = 1 Zn2+ ion and 1/2(4) = 2 Al3+ ions present along with the 4
S2 ions. The formula is ZnAl2S4.
66.
The two-dimension unit cell is:
Assuming the anions A are the larger circles, there are 4 anions completely in the unit cell.
The corner cations (smaller circles) are shared by 4 different unit cells. Therefore, there is 1
cation in the middle of the unit cell plus 1/4(4) = 1 net cation from the corners. Each unit cell
has 2 cations and 4 anions. The empirical formula is MA2.
350
CHAPTER 10
LIQUIDS AND SOLIDS
67.
8 F ions at corners × 1/8 F/corner = 1 F ion per unit cell; Because there is one cubic hole
per cubic unit cell, there is a 2:1 ratio of F ions to metal ions in the crystal if only ½ of the
body centers are filled with the metal ions. The formula is MF2 where M2+ is the metal ion.
68.
Mn ions at 8 corners: 8(1/8) = 1 Mn ion; F ions at 12 edges: 12(1/4) = 3 F ions
Formula is MnF3. Assuming fluoride is -1 charged, the charge on Mn is +3.
69.
From Fig. 10.37, MgO has the NaCl structure containing 4 Mg2+ ions and 4 O2 ions per facecentered unit cell.
4 MgO formula units ×
1 mol MgO
40.31 g MgO
= 2.678 × 10 22 g MgO

23
1 mol MgO
6.022  10 atoms
Volume of unit cell = 2.678 × 10 22 g MgO ×
1 cm3
= 7.48 × 10 23 cm3
3.58 g
Volume of unit cell = l3, l = cube edge length; l = (7.48 × 10 23 cm3)1/3 = 4.21 × 10 8 cm
For a face-centered unit cell, the O2 ions touch along the face diagonal:
2 l  4 rO 2 , rO 2 
2  4.21  10 8 cm
= 1.49 × 10 8 cm
4
The cube edge length goes through two radii of the O2 anions and the diameter of the Mg2+
cation. So:
l = 2 rO 2  2 rMg 2 , 4.21 × 10 8 cm = 2(1.49 × 10 8 cm) + 2 rMg 2 , rMg 2 = 6.15 × 10 9 cm
70.
CsCl is a simple cubic array of Cl ions with Cs+ in the middle of each unit cell. There is one
Cs+ and one Cl ion in each unit cell. Cs+ and Cl ions touch along the body diagonal.
body diagonal = 2 rCs   2 rCl   3 l , l = length of cube edge
In each unit cell:

  168 .4 g CsCl 
1 mol CsCl
 

mass = 1 CsCl formula unit 
23

 6.022  10 formula units   mol CsCl 
= 2.796 × 10 22 g
1 cm3
volume = l 3 = 2.796 × 10 22 g CsCl ×
= 7.04 × 10 23 cm3
3.97 g CsCl
l 3 = 7.04 × 10 23 cm3, l = 4.13 × 10 8 cm = 413 pm = length of cube edge
2 rCs   2 rCl   3 l  3(413 pm) = 715 pm
CHAPTER 10
LIQUIDS AND SOLIDS
351
The distance between ion centers = rCs   rCl  = 715 pm/2 = 358 pm
From ionic radii: rCs  = 169 pm and rCl  = 181 pm; rCs   rCl  = 169 + 181 = 350. pm
The actual distance is 8 pm (2.3%) greater than that calculated from values of ionic radii.
71.
72.
73.
a. CO2: molecular
b. SiO2: network
c. Si: atomic, network
d. CH4: molecular
e. Ru: atomic, metallic
f.
I2: molecular
g. KBr: ionic
h. H2O: molecular
i.
NaOH: ionic
j.
k. CaCO3: ionic
l.
PH3: molecular
U: atomic, metallic
a. diamond: atomic, network
b. PH3: molecular
c. H2: molecular
d. Mg: atomic, metallic
e. KCl: ionic
f.
quartz: network
g. NH4NO3: ionic
h. SF2: molecular
i.
Ar: atomic, group 8A
j.
k. C6H12O6: molecular
Cu: atomic, metallic
a. The unit cell consists of Ni at the cube corners and Ti at the body center, or Ti at the cube
corners and Ni at the body center.
b. 8 × 1/8 = 1 atom from corners + 1 atom at body center; Empirical formula = NiTi
c. Both have a coordination number of 8 (both are surrounded by 8 atoms).
74.
8 corners ×
1 / 8 Xe
1/ 4 F
+ 1 Xe inside cell = 2 Xe; 8 edges ×
+ 2 F inside cell = 4 F
edge
corner
Empirical formula is XeF2. This is also the molecular formula.
75.
Structure 1
8 corners ×
6 faces ×
1 / 8 Ca
= 1 Ca atom
corner
1/ 2 O
= 3 O atoms
face
1 Ti at body center. Formula = CaTiO3
Structure 2
8 corners ×
1/ 8 Ti
= 1 Ti atom
corner
12 edges ×
1/ 4 O
= 3 O atoms
corner
1 Ca at body center. Formula = CaTiO3
In the extended lattice of both structures, each Ti atom is surrounded by six O atoms.
76.
With a cubic closest packed array of oxygen ions, we have 4 O2 ions per unit cell. We need
to balance the total 8 charge of the anions with a +8 charge from the Al3+ and Mg2+ cations.
352
CHAPTER 10
LIQUIDS AND SOLIDS
The only combination of ions that gives a +8 charge is 2 Al3+ ions and 1 Mg2+ ion. The
formula is Al2MgO4.
There are an equal number of octahedral holes as anions (4) in a cubic closest packed array,
and twice the number of tetrahedral holes as anions in a cubic closest packed array. For the
stoichiometry to work out, we need 2 Al3+ and 1 Mg2+ per unit cell. Hence, one-half of the
octahedral holes are filled with Al3+ ions and one-eighth of the tetrahedral holes are filled
with Mg2+ ions.
77.
a. Y: 1 Y in center; Ba: 2 Ba in center
Cu: 8 corners ×
O: 20 edges ×
1 / 8 Cu
1 / 4 Cu
= 1 Cu, 8 edges ×
= 2 Cu, total = 3 Cu atoms
corner
edge
1/ 2 O
1/ 4 O
= 5 oxygen, 8 faces ×
= 4 oxygen, total = 9 O atoms
face
edge
Formula: YBa2Cu3O9
b. The structure of this superconductor material follows the second perovskite structure
described in Exercise 10.75. The YBa2Cu3O9 structure is three of these cubic perovskite
unit cells stacked on top of each other. The oxygen atoms are in the same places, Cu
takes the place of Ti, two of the calcium atoms are replaced by two barium atoms, and
one Ca is replaced by Y.
c. Y, Ba, and Cu are the same. Some oxygen atoms are missing.
12 edges ×
1/ 2 O
1/ 4 O
= 3 O, 8 faces ×
= 4 O, total = 7 O atoms
face
edge
Superconductor formula is YBa2Cu3O7.
78.
a. Structure (a):
1/ 8 Tl
= 1 Tl
corner
Ba: 2 Ba inside unit cell; Tl: 8 corners ×
1 / 4 Cu
= 1 Cu
edge
1/ 2 O
1/ 4 O
O: 6 faces ×
+ 8 edges ×
= 5 O; Formula = TlBa2CuO5
face
edge
Cu: 4 edges ×
Structure (b):
Tl and Ba are the same as in structure (a).
Ca: 1 Ca inside unit cell; Cu: 8 edges ×
O: 10 faces ×
1 / 4 Cu
= 2 Cu
edge
1/ 2 O
1/ 4 O
+ 8 edges ×
= 7 O; Formula = TlBa2CaCu2O7
face
edge
CHAPTER 10
LIQUIDS AND SOLIDS
353
Structure (c):
Tl and Ba are the same, and two Ca are located inside the unit cell.
Cu: 12 edges ×
1/ 2 O
1 / 4 Cu
1/ 4 O
= 3 Cu; O: 14 faces ×
+ 8 edges ×
=9O
edge
face
edge
Formula: TlBa2Ca2Cu3O9
Structure (d):
Following similar calculations, formula = TlBa2Ca3Cu4O11
b. Structure (a) has one planar sheet of Cu and O atoms, and the number increases by one
for each of the remaining structures. The order of superconductivity temperature from
lowest to highest temperature is: (a) < (b) < (c) < (d).
c. TlBa2CuO5: 3 + 2(2) + x + 5(2) = 0, x = +3
Only Cu3+ is present in each formula unit.
TlBa2CaCu2O7: 3 + 2(2) + 2 + 2(x) + 7(2) = 0, x = +5/2
Each formula unit contains 1 Cu2+ and 1 Cu3+.
TlBa2Ca2Cu3O9: 3 + 2(2) + 2(2) + 3(x) + 9(2) = 0, x = +7/3
Each formula unit contains 2 Cu2+ and 1 Cu3+.
TlBa2Ca3Cu4O11: 3 + 2(2) + 3(2) + 4(x) + 11(2) = 0, x = +9/4
Each formula unit contains 3 Cu2+ and 1 Cu3+.
d. This superconductor material achieves variable copper oxidation states by varying the
numbers of Ca, Cu and O in each unit cell. The mixtures of copper oxidation states are
discussed above. The superconductor material in Exercise 10.77 achieves variable
copper oxidation states by omitting oxygen at various sites in the lattice.
Phase Changes and Phase Diagrams
79.
If we graph ln Pvap vs 1/T, the slope of the resulting straight line will be ΔHvap/R.
Pvap
ln Pvap
T (Li)
1 torr
10.
100.
400.
760.
0
2.3
4.61
5.99
6.63
1023 K
1163
1353
1513
1583
1/T
9.775 × 104 K1
8.598 × 104
7.391 × 104
6.609 × 104
6.317 × 104
T (Mg)
893 K
1013
1173
1313
1383
1/T
11.2 × 104 K1
9.872 × 104
8.525 × 104
7.616 × 104
7.231 × 104
354
CHAPTER 10
LIQUIDS AND SOLIDS
For Li:
We get the slope by taking two points (x, y) that are on the line we draw. For a line:
slope =
y y
Δy
 2 1
Δx
x 2  x1
or we can determine the straight line equation using a calculator. The general straight
line equation is y = mx + b where m = slope and b = y-intercept.
The equation of the Li line is: ln Pvap = 1.90 × 104(1/T) + 18.6, slope = 1.90 × 104 K
Slope = ΔHvap/R, ΔHvap = slope × R = 1.90 × 104 K × 8.3145 J/Kmol
ΔHvap = 1.58 × 105 J/mol = 158 kJ/mol
For Mg:
The equation of the line is: ln Pvap = 1.67 × 104(1/T) + 18.7, slope = 1.67 × 104 K
ΔHvap = -slope × R = 1.67 × 104 K × 8.3145 J/Kmol,
ΔHvap = 1.39 × 105 J/mol = 139 kJ/mol
The bonding is stronger in Li since ΔHvap is larger for Li.
80.
We graph ln Pvap vs 1/T. The slope of the line equals ΔHvap/R.
CHAPTER 10
LIQUIDS AND SOLIDS
103/T (K1) Pvap (torr)
T(K)
273
283
293
303
313
323
353
slope =
3.66
3.53
3.41
3.30
3.19
3.10
2.83
14.4
26.6
47.9
81.3
133
208
670.
355
ln Pvap
2.67
3.28
3.87
4.40
4.89
5.34
6.51
6.6  2.5
(2.80  10
-4600 K =
3
 ΔH vap
R
= 4600 K
 3.70  10 3 ) K 1

 ΔH vap
8.3145 J / K  mol
, ΔHvap = 38,000 J/mol = 38 kJ/mol
To determine the normal boiling point, we can use the following formula:
 P  ΔH vap
 1
1
=   
ln  1  =
R
 P2 
 T2 T1 
At the normal boiling point, the vapor pressure equals 1.00 atm or 760. torr. At 273 K, the
vapor pressure is 14.4. torr (from data in the problem).
38,000 J / mol  1
1 
 14.4 
 
 , 3.97 = 4.6 × 103 (1/T2  3.66 × 103)
ln 
 =
 760 .  8.3145 J / K  mol  T2 273 K 
8.6 × 104 + 3.66 × 103 = 1/T2 = 2.80 × 103, T2 = 357 K = normal boiling point
81.
At 100.°C (373 K), the vapor pressure of H2O is 1.00 atm = 760. torr.
For water, ΔHvap = 40.7 kJ/mol.
P
ln  1
 P2
 ΔH vap  1
P
1 


 or ln  2




P
R  T2 T1 

 1
 ΔH vap  1
1 






R  T1 T2 

 520 . torr  40.7  10 3 J / mol  1
1 

 , 7.75 × 105 =
 =
ln 

760
.
torr
8
.
3145
J
/
K

mol
373
K
T


2 

7.75 × 105 = 2.68 × 103 
82.
 1
1 



 373 K T2 
1
1
1
= 2.76 × 103, T2 =
= 362 K or 89°C
,
2.76  10  3
T2 T2
40.7  10 3 J / mol  1
1 
 P 

 , ln P2 = 5.27, P2 = e5.27 = 194 atm

ln  2  =
8
.
3145
J
/
K

mol
373
K
623
K
1
.
00




356
83.
CHAPTER 10
P
ln  1
 P2
LIQUIDS AND SOLIDS
ΔH vap
 ΔH vap  1
 836 torr 
 1
1 
1 


, ln 
 







R  T2 T1 
 213 torr  8.3145 J / K  mol  313 K 353 K 

Solving: Hvap = 3.1 × 104 J/mol; For the normal boiling point, P = 1.00 atm = 760. torr.
 760 . torr 
3.1  10 4 J / mol  1
1
1
1

 =
= 3.4 × 10 4
ln 
  ,

213
torr
8
.
3145
J
/
K

mol
313
K
T
313
T


1
1

T1 = 350. K = 77C; The normal boiling point of CCl4 is 77C.
84.
P
ln  1
 P2
 ΔH vap  1
1 






R  T2 T1 

P1 = 760. torr, T1 = 56.5°C + 273.2 = 329.7 K; P2 = 630. torr, T2 = ?
 760 .
ln 
 630 .
 32.0  10 3 J / mol  1
1

 =

8.3145 J / K  mol  T2
329 .7


 , 0.188 = 3.85 × 103

 1


 3.033  10 3 
 T2

1
1
3
5
 3.033 × 10 = 4.88 × 10 ,
= 3.082 × 10 3 , T2 = 324.5 K = 51.3°C
T2
T2
 630 . torr
ln 
 P2
 32.0  10 3 J / mol  1
1
 =


 8.3145 J / K  mol  298.2 324.5
ln P2 = 5.40, P2 = e5.40 = 221 torr
85.

 , ln 630.  ln P2 = 1.05

CHAPTER 10
86.
LIQUIDS AND SOLIDS
357
X(g, 100.°C) → X(g, 75°C), ΔT = -25°C
q1 = sgas × m × ΔT =
1.0 J
× 250. g × (25°C) = 6300 J = 6.3 kJ
gC
X(g, 75°C) → X(l, 75°C), q2 = 250. g ×
X(l, 75°C) → X(l, -15°C), q3 =
1 mol  20. kJ
= 67 kJ

75.0 g
mol
2.5 J
× 250. g × (-90.°C) = -56,000 J = 56 kJ
gC
X(l, 15°C) → X(s, 15°C), q4 = 250. g ×
X(s, 15°C) → X(s, 50.°C), q5 =
1 mol  5.0 kJ
= 17 kJ

75.0 g
mol
3.0 J
× 250. g × (35°C) = 26,000 J = 26 kJ
gC
qtotal = q1 + q2 + q3 + q4 + q5 = -6.3  67 - 56  17  26 = 172 kJ
87.
H2O(s, 20.°C) → H2O(s, 0°C), ΔT = 20.°C
q1 = sice × m × ΔT =
2.03 J
× 5.00 × 102 g × 20.°C = 2.0 × 104 J = 20. kJ
gC
H2O(s, 0°C) → H2O(l, 0°C), q2 = 5.00 × 102 g H2O ×
H2O(l, 0°C) → H2O(l, 100.°C), q3 =
1 mol 6.02 kJ
= 167 kJ

18.02 g
mol
4.2 J
× 5.00 × 102 g × 100.°C = 2.1 × 105 J = 210 kJ
gC
H2O(l, 100.°C) → H2O(g, 100.°C), q4 = 5.00 × 102 g ×
H2O(g, 100.°C) → H2O(g, 250.°C), q5 =
1 mol 40.7 kJ
= 1130 kJ

18.02 g
mol
2.0 J
× 5.00 × 102 g × 150.°C = 1.5 × 105 J
gC
= 150 kJ
qtotal = q1 + q2 + q3 + q4 + q5 = 20. + 167 + 210 + 1130 + 150 = 1680 kJ
88.
H2O(g, 125°C) → H2O(g, 100.°C), q1 = 2.0 J/g °C × 75.0 g × (25°C) = -3800 J = 3.8 kJ
H2O(g, 100.°C) → H2O(l, 100.°C), q2 = 75.0 g ×
1 mol  40.7 kJ
= 169 kJ

18.02 g
mol
H2O(l, 100.°C) → H2O(l, 0°C), q3 = 4.2 J/g °C × 75.0 g × (-100.°C) = 32,000 J = 32 kJ
To convert H2O(g) at 125°C to H2O(l) at 0°C requires (3.8 kJ  169 kJ  32 kJ =) 205 kJ of
heat removed. To convert from H2O(l) at 0°C to H2O(s) at 0°C requires:
358
CHAPTER 10
LIQUIDS AND SOLIDS
1 mol  6.02 kJ
= 25 kJ

18.02 g
mol
This amount of energy puts us over the -215 kJ limit (-205 kJ - 25 kJ = -230. kJ). Therefore,
a mixture of H2O(s) and H2O(l) will be present at 0°C when 215 kJ of heat are removed from
the gas sample.
q4 = 75.0 g ×
89.
Total mass H2O = 18 cubes ×
30.0 g
1 mol H 2O
= 540. g; 540. g H2O ×
= 30.0 mol H2O
cube
18.02 g
Heat removed to produce ice at -5.0°C:
6.02  10 3 J
4.18 J
2.03 J
× 540. g × 22.0 °C +
× 30.0 mol +
× 540. g × 5.0 °C
mol
gC
gC
4.97 × 104 J + 1.81 × 105 J + 5.5 × 103 J = 2.36 × 105 J
1 g CF2Cl2
= 1.49 × 103 g CF2Cl2 must be vaporized.
158 J
1 mol
368 kJ
Heat released = 0.250 g Na ×
= 2.00 kJ

22.99 g 2 mol
2.36 × 105 J ×
90.
To melt 50.0 g of ice requires: 50.0 g ice ×
1 mol H 2 O 6.02 kJ
= 16.7 kJ

18.02 g
mol
The reaction doesn't release enough heat to melt all of the ice. The temperature will remain at
0°C.
91.
A: solid
B: liquid
D: solid + vapor
E: solid + liquid + vapor
F: liquid + vapor
G: liquid + vapor
triple point: E
C: vapor
H: vapor
critical point: G
normal freezing point: temperature at which solid - liquid line is at 1.0 atm (see plot below).
normal boiling point: temperature at which liquid - vapor line is at 1.0 atm (see plot below).
1.0
atm
nfp
nbp
Since the solid-liquid line has a positive slope, the solid phase is denser than the liquid phase.
CHAPTER 10
92.
LIQUIDS AND SOLIDS
359
a. 3
b. Triple point at 95.31°C: rhombic, monoclinic, gas
Triple point at 115.18°C: monoclinic, liquid, gas
Triple point at 153°C: rhombic, monoclinic, liquid
c. From the phase diagram, the monoclinic solid phase is stable at T = 100°C and P = 1 atm.
d. Normal melting point = 115.21°C; normal boiling point = 444.6°C; The normal melting
and boiling points occur at P = 1.0 atm.
e. Rhombic is the densest phase since the rhombic-monoclinic equilibrium line has a positive slope and since the solid-liquid lines also have positive slopes.
No; P = 1.0 × 10 5 atm is at a pressure somewhere between the 95.31°C and 115.18°C
triple points. At this pressure, the rhombic and gas phases are never in equilibrium with
each other, so rhombic sulfur cannot sublime at P = 1.0 × 10 5 atm. However, monoclinic
sulfur can sublime at this pressure.
g. From the phase diagram, we would start off with gaseous sulfur. At 100°C and ~1 × 10 5
atm, S(g) would convert to the solid monoclinic form of sulfur. Finally at 100°C and
some large pressure less than 1420 atm, S(s, monoclinic) would convert to the solid
rhombic form of sulfur. Summarizing, the phase changes are S(g) ➔ S(monoclinic) ➔
S(rhombic).
f.
93.
a. two
b. Higher pressure triple point: graphite, diamond and liquid; Lower pressure triple point:
graphite, liquid and vapor
c. It is converted to diamond (the more dense solid form).
d. Diamond is more dense, which is why graphite can be converted to diamond by applying
pressure.
94.
The following sketch of the Br2 phase diagram is not to scale. Since the triple point of Br2 is
at a temperature below the freezing point of Br2, the slope of the solid-liquid line is positive.
100
Solid
Pressure (atm)
Liquid
1
0.05
Gas
-7.3 -7.2
59
Temperature ( oC)
320
360
CHAPTER 10
LIQUIDS AND SOLIDS
The positive slopes of all the lines indicate that Br2(s) is more dense than Br2(l) which is more
dense than Br2(g). At room temperature (~22°C) and 1 atm, Br 2(l) is the stable phase. Br2(l)
cannot exist at a temperature below the triple point temperature of 7.3°C and at a temperature above the critical point temperature of 320°C. The phase changes that occur as
temperature is increased at 0.10 atm are solid ➔ liquid ➔ gas.
95.
Because the density of the liquid phase is greater than the density of the solid phase, the slope
of the solid-liquid boundary line is negative (as in H2O). With a negative slope, the melting
points increase with a decrease in pressure so the normal melting point of X should be greater
than 225C.
96.
760
s
l
P
(torr)
g
280
-121
-112
-107
T(C)
From the three points given, the slope of the s-l boundary line is positive so Xe(s) is more
dense than Xe(l). Also, the positive slope of this line tells us that the melting point of Xe
increases as pressure increases. The same direct relationship exists for the boiling point of Xe
as the l-g boundary line also has a positive slope.
Additional Exercises
97.
Chalk is composed of the ionic compound calcium carbonate (CaCO3). The electrostatic
forces in ionic compounds are much stronger than the intermolecular forces in covalent
compounds. Therefore, CaCO3 should have a much higher boiling point than the covalent
compounds found in motor oil and in H2O. Motor oil is composed of nonpolar CC and CH
bonds. The intermolecular forces in motor oil are therefore London dispersion forces. We
generally consider these forces to be weak. However, with compounds that have large molar
masses, these London dispersion forces add up significantly and can overtake the relatively
strong hydrogen bonding interactions in water.
CHAPTER 10
98.
LIQUIDS AND SOLIDS
Benzene
361
Naphthalene
H
H
H
H
H
H
H
H
H
H
H
H
H
LD forces only
H
LD forces only
Note: London dispersion forces in molecules like benzene and naphthalene are fairly large.
The molecules are flat, and there is efficient surface area contact among molecules. Large
surface area contact leads to stronger London dispersion forces.
Cl
Carbon tetrachloride (CCl4) has polar bonds but is a nonpolar
molecule. CCl4 only has LD forces.
C
Cl
Cl
Cl
In terms of size and shape: CCl4 < C6H6 < C10H8
The strengths of the LD forces are proportional to size and are related to shape. Although
CCl4 is fairly large, its overall spherical shape gives rise to relatively weak LD forces as
compared to flat molecules like benzene and naphthalene. The physical properties given in
the problem are consistent with the order listed above. Each of the physical properties will
increase with an increase in intermolecular forces.
Acetone
Acetic Acid
O
O
H 3C
C
CH 3
LD, dipole
H 3C
C
O
H
LD, dipole, H bonding
362
CHAPTER 10
LIQUIDS AND SOLIDS
Benzoic acid
O
C
O
H
LD, dipole, H bonding
We would predict the strength of intermolecular forces for the last three molecules to be:
acetone < acetic acid < benzoic acid
polar
H bonding
H bonding, but large LD forces because of greater size and
shape.
This ordering is consistent with the values given for bp, mp, and ΔHvap.
The overall order of the strengths of intermolecular forces based on physical properties are:
acetone < CCl4 < C6H6 < acetic acid < naphthalene < benzoic acid
The order seems reasonable except for acetone and naphthalene. Since acetone is polar, we
would not expect it to boil at the lowest temperature. However, in terms of size and shape,
acetone is the smallest molecule, and the LD forces in acetone must be very small compared
to the other molecules. Naphthalene must have very strong LD forces because of its size and
flat shape.
99.
At any temperature, the plot tells us that substance A has a higher vapor pressure than substance B, with substance C having the lowest vapor pressure. Therefore, the substance with
the weakest intermolecular forces is A, and the substance with the strongest intermolecular
forces is C.
NH3 can form hydrogen bonding interactions while the others cannot. Substance C is NH3.
The other two are nonpolar compounds with only London dispersion forces. Since CH4 is
smaller than SiH4, CH4 will have weaker LD forces and is substance A. Therefore, substance
B is SiH4.
100.
As the electronegativity of the atoms covalently bonded to H increases, the strength of the
hydrogen bonding interaction increases.
N ⋅⋅⋅ H‒N < N ⋅⋅⋅ H‒ O < O ⋅⋅⋅ H‒O < O ⋅⋅⋅ H‒F < F ⋅⋅⋅ H‒F
weakest
101.
strongest
If TiO2 conducts electricity as a liquid, then it is an ionic solid; if not, then TiO2 is a network
solid.
CHAPTER 10
LIQUIDS AND SOLIDS
363
102.
One B atom and one N atom together have the same number of electrons as two C atoms. The
description of physical properties sounds a lot like the properties of graphite and diamond, the
two solid forms of carbon. The two forms of BN have structures similar to graphite and
diamond.
103.
B2H6: This compound contains only nonmetals so it is probably a molecular solid with
covalent bonding. The low boiling point confirms this.
104.
SiO2:
This is the empirical formula for quartz, which is a network solid.
CsI:
This is a metal bonded to a nonmetal, which generally form ionic solids. The
electrical conductivity in aqueous solution confirms this.
W:
Tungsten is a metallic solid as the conductivity data confirms.
In order to set up an equation, we need to know what phase exists at the final temperature. To
heat 20.0 g of ice from 10.0°C to 0.0°C requires:
2.03 J
× 20.0 g × 10.0C = 406 J
gC
To convert ice to water at 0.0°C requires:
q=
q = 20.0 g ×
1 mol 6.02 kJ
= 6.68 kJ = 6680 J

18.02
mol
To chill 100.0 g of water from 80.0°C to 0.0° requires:
4.18 J
× 100.0 g × 80.0°C = 33,400 J of heat removed
gC
From the heat values above, the liquid phase exists once the final temperature is reached (a
lot more heat is lost when the 100.0 g of water is cooled to 0.0°C than the heat required to
convert the ice into water). To calculate the final temperature, we will equate the heat gain by
the ice to the heat loss by the water. We will keep all quantities positive in order to avoid sign
errors. The heat gain by the ice will be the 406 J required to convert the ice to 0.0°C plus the
6680 J required to convert the ice at 0.0°C into water at 0.0°C plus the heat required to raise
the temperature from 0.0°C to the final temperature.
q=
heat gain by ice = 406 J + 6680 J +
heat loss by water =
4.18 J
× 20.0 g × (Tf  0.0°C) = 7.09 × 103 + 83.6 Tf
gC
4.18 J
× 100.0 g × (80.0°C  Tf) = 3.34 × 104  418 Tf
gC
Solving for the final temperature:
7.09 × 103 + 83.6 Tf = 3.34 × 104 - 418 Tf, 502 Tf = 2.63 × 104, Tf = 52.4°C
105.
1.00 lb ×
454 g
5
= 454 g H2O; A change of 1.00°F is equal to a change of C .
lb
9
364
CHAPTER 10
The amount of heat in J in 1 Btu is:
LIQUIDS AND SOLIDS
4.18 J
5
 454 g  C = 1.05 × 103 J = 1.05 kJ
gC
9
It takes 40.7 kJ to vaporize 1 mol H2O (ΔHvap). Combining these:
1.00  10 4 Bu 1.05 kJ 1 mol H 2 O


= 258 mol/hr
hr
Btu
40.7 kJ
or:
106.
258 mol 18.02 g H 2O
= 4650 g/hr = 4.65 kg/hr

hr
mol
The critical temperature is the temperature above which the vapor cannot be liquefied no
matter what pressure is applied. Since N2 has a critical temperature below room temperature
(~22°C), it cannot be liquefied at room temperature. NH3, with a critical temperature above
room temperature, can be liquefied at room temperature.
Challenge Problems
107.
H = qp = 30.79 kJ; E = qp + w, w = PV
101 .3 J
L atm
w = PV = 1.00 atm (28.90 L) = 28.9 L atm ×
= 2930 J
E = 30.79 kJ + (2.93 kJ) = 27.86 kJ
108.
XeCl2F2, 8 + 2(7) + 2(7) = 36 e
Cl
F
F
or
Xe
Cl
Cl
F
Xe
F
polar (bond dipoles do
not each other cancel)
Cl
nonpolar (bond dipoles
cancel each other)
These are two possible square planar molecular structures for XeCl2F2. One structure has the
Cl atoms 90 apart, the other has the Cl atoms 180 apart. The structure with the Cl atoms
90 apart is polar; the other structure is nonpolar. The polar structure will have additional
dipole forces so it has the stronger intermolecular forces and is the liquid. The gas form of
XeCl2F2 is the nonpolar form having the Cl atoms 180 apart.
109.
A single hydrogen bond in H2O has a strength of 21 kJ/mol. Each H2O molecule forms two
H bonds. Thus, it should take 42 kJ/mol of energy to break all of the H bonds in water.
Consider the phase transitions:
solid
6.0 kJ
liquid
40.7 kJ
vapor
Hsub = fusHvap
CHAPTER 10
LIQUIDS AND SOLIDS
365
It takes a total of 46.7 kJ/mol to convert solid H2O to vapor (ΔHsub). This would be the
amount of energy necessary to disrupt all of the intermolecular forces in ice. Thus,
(42 ÷ 46.7) × 100 = 90% of the attraction in ice can be attributed to H bonding.
110.
Both molecules are capable of H bonding. However, in oil of wintergreen the hydrogen
bonding is intramolecular (within each molecule).
O
CH 3
C
O
H
O
In methyl-4-hydroxybenzoate, the H bonding is intermolecular, resulting in stronger forces
between molecules and a higher melting point.
111.
NaCl, MgCl2, NaF, MgF2 AlF3 all have very high melting points indicative of strong intermolecular forces. They are all ionic solids. SiCl4, SiF4, F2, Cl2, PF5 and SF6 are nonpolar
covalent molecules. Only LD forces are present. PCl3 and SCl2 are polar molecules. LD
forces and dipole forces are present. In these 8 molecular substances, the intermolecular
forces are weak and the melting points low. AlCl3 doesn't seem to fit in as well. From the
melting point, there are much stronger forces present than in the nonmetal halides, but they
aren't as strong as we would expect for an ionic solid. AlCl3 illustrates a gradual transition
from ionic to covalent bonding, from an ionic solid to discrete molecules.
112.
a. The NaCl unit cell has a face centered cubic arrangement of the anions with cations in the
octahedral holes. There are 4 NaCl formula units per unit cell and, since there is a 1:1
ratio of cations to anions in MnO, there would be 4 MnO formula units per unit cell,
assuming an NaCl type structure. The CsCl unit cell has a simple cubic structure of
anions with the cations in the cubic holes. There is one CsCl formula unit per unit cell, so
there would be one MnO formula unit per unit cell if a CsCl structure is observed.
molecules MnO
5.28 g MnO 1 mol MnO
 (4.47  10 8 cm)3 

unit cell
70.94 g MnO
cm3

6.022  10 23 molecules MnO
= 4.00 molecules MnO
mol MnO 4
From the calculation, MnO crystallizes in the NaCl type structure.
b. From the NaCl structure and assuming the ions touch each other, then ℓ = cube edge
length = 2rMn 2   2rO2  .
ℓ = 4.47 × 10 8 cm = 2 rMn 2 + 2(1.40 × 10 8 cm), rMn 2 = 8.4 × 10 9 cm = 84 pm
366
113.
CHAPTER 10
Out of 100.00 g: 28.31 g O ×
LIQUIDS AND SOLIDS
1 mol
1 mol
= 1.769 mol O; 71.69 g Ti ×
= 1.497 mol Ti
16.00 g
47.88 g
1.497
1.769
= 1.182;
= 0.8462; The formula is TiO1.182 or Ti0.8462O.
1.769
1.497
For Ti0.8462O, let x = Ti2+ per mol O2- and y = Ti3+ per mol O2-. Setting up two equations and
solving:
x + y = 0.8462 (mass balance) and 2x + 3y = 2 (charge balance); 2x + 3(0.8462  x) = 2
x = 0.539 mol Ti2+/mol O2- and y = 0.307 mol Ti3+/mol O2-
0.539
× 100 = 63.7% of the titanium ions are Ti2+ and 36.3% are Ti3+ (a 1.75:1 ion ratio).
0.8462
114.
First we need to get the empirical formula of spinel. Assume 100.0 g of spinel.
37.9 g Al ×
1 mol Al
= 1.40 mol Al
26.98 g Al
The mole ratios are 2:1:4.
17.1 g Mg ×
1 mol Mg
= 0.703 mol Mg
24.31 g Mg
Empirical Formula = Al2MgO4
1 mol O
45.0 g O ×
= 2.81 mol O
16.00 g O
Assume each unit cell contains an integral value (n) of Al2MgO4 formula units. Each
Al2MgO4 formula unit has a mass of: 24.31 + 2(26.98) + 4(16.00) = 142.27 g/mol
1 mol
142 .27 g

23
mol
3.57 g
6.022  10 atoms
, Solving: n = 8.00

8
3
(8.09  10 cm)
cm3
n formula units 
density =
Each unit cell has 8 formula units of Al2MgO4 or 16 Al3+, 8 Mg2+ and 32 O2 ions.
115.
volume Cu
density Mn massMn  volume Cu
massMn



density Cu volume Mn  massCu
massCu
volume Mn
The type of cubic cell formed is not important; only that Cu and Mn crystallize in the same
type of cubic unit cell is important. Each cubic unit cell has a specific relationship between
the cube edge length, l, and the radius, r. In all cases l  r. Therefore, V  l3  r3. For the
mass ratio, we can use the molar masses of Mn and Cu since each unit cell must contain the
same number of Mn and Cu atoms. Solving:
CHAPTER 10
LIQUIDS AND SOLIDS
367
volume Cu 54.94 g / mol
(rCu ) 3
density Mn massMn




density Cu
massCu volume Mn 63.55 g / mol (1.056 rCu ) 3
3
density Mn
 1 
= 0.8645 × 
 = 0.7341
density Cu
 1.056 
densityMn = 0.7341 × densityCu = 0.7341 × 8.96 g/cm3 = 6.58 g/cm3
116.
a. The arrangement of the layers are:
Layer 1
Layer 2
Layer 3
Layer 4
A total of 20 cannon balls will be needed.
b. The layering alternates abcabc which is cubic closest packing.
c. tetrahedron
117.
a
P
c
b
As P is lowered, we go from a to b on the phase
diagram. The water boils. The boiling of water is
endothermic and the water is cooled (b →c),
forming some ice. If the pump is left on, the ice
will sublime until none is left. This is the basis of
freeze drying.
T
118.
w = PΔV;
V373 =
Assume a constant P of 1.00 atm.
nRT 1.00 (0.8206 ) (373)
= 30.6 L for one mol of water vapor

P
1.00
Since the density of H2O(l) is 1.00 g/cm3, 1.00 mol of H2O(l) occupies 18.0 cm3 or 0.0180 L.
w = 1.00 atm (30.6 L  0.0180 L) = 30.6 L atm
w = 30.6 L atm × 101.3 J/Latm = 3.10 × 103 J = 3.10 kJ
368
CHAPTER 10
LIQUIDS AND SOLIDS
ΔE = q + w = 40.7 kJ  3.10 kJ = 37.6 kJ
37.6
× 100 = 92.4% of the energy goes to increase the internal energy of the water.
40.7
The remainder of the energy (7.6%) goes to do work against the atmosphere.
119.
For a cube: (body diagonal)2 = (face diagonal)2 + (cube edge length)2
In a simple cubic structure, the atoms touch on cube edge so the cube edge = 2 r where r
= radius of sphere.
2r
2r
face diagonal =
body diagonal =
(2 r ) 2  (2 r ) 2 =
4 r2  4 r2 = r 8 = 2 2 r
(2 2 r ) 2  (2 r ) 2 = 12 r 2 = 2 3 r
The diameter of the hole = body diagonal  2 radius of atoms at corners.
diameter = 2 3 r  2 r; Thus, the radius of the hole is:
4  2 3  2  
The volume of the hole is: π 
r
3 
2  
120.
2 3 r  2 r  2 3  2 

r
2
2


3
C2H6O, 2(4) + 6(1) + 6 = 20 e; Determining the two possible structures is trial and error.
Because hydrogen is always attached with a single bond, H cannot be a central atom. The two
possible structures having two bonds and two lone pairs for O are:
H
H
H
C
C
H
H
H
O
liquid (H-bonding)
H
and
H
C
H
H
O
C
H
gas (no H-bonding)
H
CHAPTER 10
LIQUIDS AND SOLIDS
369
The first structure with the –OH bond is capable of forming hydrogen bonding; the other
structure is not. Therefore, the liquid (which has the stronger intermolecular forces) is the
first structure and the gas is the second structure.
Integrative Problems
121.
19.0 g
= 144 g/mol
0.132 mol
molar mass XY =
X: [Kr] 5s24d10; This is cadmium, Cd.
molar mass Y = 144 – 112.4 = 32 g/mol; Y is sulfur, S.
The semiconductor is CdS. The dopant has the electron configuration of bromine, Br.
Because Br has one more valence electron than S, doping with Br will produce an n-type
semiconductor.
122.
Assuming 100.00 g of MO2:
23.72 g O ×
1 mol O
= 1.483 mol O
16.00 g O
1.483 mol O ×
1 mol M
= 0.7415 mol M
2 mol O
100.00 g – 23.72 g = 76.28 g M
molar mass M =
76.28 g
= 102.9 g/mol
0.7415 mol
From the periodic table element M is rhodium, Rh.
The unit cell for cubic closest packing is face-centered cubic (4 atoms/unit cell). The atoms
for fcc are assumed to touch along the face diagonal of the cube so face diagonal = 4 r. The
distance between the centers of touching Rh atoms will be the distance of 2 r where r = radius
of Rh atom.
face diagonal =
2 l, where l = cube edge
face diagonal = 4 r = 2 × 269.0 × 10 12 m = 5.380 × 10 10 m
2 l = 4 r = 5.38 × 10 10 m, l =
4 atoms Rh 
density =
5.38  10 10 m
2
= 3.804 × 10 10 m = 3.804 × 10 8 cm
1 mol Rh
102 .9 g Rh

23
mol Rh
6.0221  10 atoms
= 12.42 g/cm3
8
3
(3.804  10 cm)
370
123.
CHAPTER 10
P
ln  1
 P2
LIQUIDS AND SOLIDS
 ΔH vap  1
1 
296 J 200 .6 g


; ΔH vap 
= 5.94 × 104 J/mol Hg





R
g
mol
T
T
1 

 2
 2.56  10 3 torr 
5.94  10 4 J / mol

ln 
 8.3145  10 4 J / K  mol
P2


 1

1



 573 K 298 .2 K 
 2.56  10 3 torr 
 = 11.5, P2 = 2.56 × 10 3 torr / e 11.5 = 253 torr
ln 

P
2



1 atm 
 253 torr 
  15.0 L
760 torr 
PV 
n=
= 0.106 mol Hg

0.08206 L atm
RT
 573 K
K mol
0.106 mol Hg ×
6.022  10 23 atoms Hg
= 6.38 × 1022 atoms Hg
mol Hg
Marathon Problem
124.
q = s × m × T; heat loss by metal = heat gain by calorimeter. The change in temperature for
the calorimeter is: T = 25.2C  0.2C – 25.0C  0.2C. Including the error limits, T
can range from 0.0C to 0.4C. Because the temperature change can be 0.0C, there is no
way that the calculated heat capacity has any meaning.
The density experiment is also not conclusive.
d=
4g
= 10 g/cm3 (1 significant figure)
3
0.42 cm
dhigh =
5g
= 12.5 g/cm3 = 10 g/cm3 to 1 sig fig
3
0.40 cm
dlow =
3g
= 7 g/cm3
3
0.44 cm
From Table 1.5, the density of copper is 8.96 g/cm3. The results from this experiment cannot
be used to distinguish between a density of 8.96 g/cm3 and 9.2 g/cm3.
The crystal structure determination is more conclusive. Assuming the metal is copper:
3
 1  10 10 cm 
  2.16  10 22 cm3
volume of unit cell = (600. pm) 

 1 pm

3
CHAPTER 10
LIQUIDS AND SOLIDS
Cu mass in unit cell = 4 atoms 
d=
371
1 mol Cu
63.55 g Cu
= 4.221 × 10 22 g Cu

23
mol Cu
6.022  10 atoms
mass
4.221  10 22 g

= 1.95 g/cm3
volume
2.16  10  22 cm3
Because the density of Cu is 8.96 g/cm3, then one can assume this metal is not copper. If the
metal is not Cu, then it must be kryptonite (as the question reads). Because we don’t know the
molar mass of kryptonite, we cannot confirm that the calculated density would be close to 9.2
g/cm3.
To improve the heat capacity experiment, a more precise balance is a must and a more precise
temperature reading is needed. Also, a larger piece of the metal should be used so that T of
the calorimeter has more significant figures. For the density experiment, we would need a
more precise balance and a more precise way to determine the volume. Again, a larger piece
of metal would help in order to insure more significant figures in the volume.