BOOK PROBLEMS C 9
3.
Well, atoms , even in a solid, are in vibrational motion at any temperature above absolute 0 K.
You know the average location of each atom precisely, as it vibrates about, but if you were to pick
out one atom and ask exactly where is it right this instant, the Heisenberg uncertainty principle
demands that the measurement you do to answer that will be subject to some error.
9.
a. Ion-ion attractions are forces of coulombic (electrostatic) attraction between two oppositely
charged ions. The potential energy of attraction varies as q2/r. That is, energy holding them
together is inversely proportional to the ion's distance apart, and directly proportional to the product
of the charges. That means that 2 ions, say Mg2+ and O2-which are highly charged and small
(therefore close together), will require more energy to break apart than a Cs+ and an I- which are
have less charge and are large (far apart). We are talking energies of for example 550 kJ/mol for Na+
and Cl- at 2.50 Å(see p R9.3)
b. Ion-dipole attractions are forces of coulombic (electrostatic) attraction between two oppositely
charged particles, one of which is an ion, and one of which is the oppositely charged end of a polar
molecule. The other end of the polar molecule repels the ion. The potential energy of attraction
varies as 1/r2. That is, energy holding them together is inversely proportional to the square of the
distance apart. That means that ion-dipole interactions are weaker than ion-ion attractions. We are
talking energies on the order of 20 kJ/mol for Na+ and H2O at 4.00 Å. (see p R9.3)
c. Dipole-dipole attractions are forces of coulombic (electrostatic) attraction between two oppositely
charged ends of polar molecules. The other ends of the polar molecules repels each other. The
potential energy of attraction varies as 1/r3. That is, energy holding them together is inversely
proportional to the cube of the distance apart, and directly proportional to the product of the charges,
and both of the charges will be less than 1. That means that dipole-dipole interactions are weaker
than ion-dipole attractions, say 2-5 kj/mol at 4.00 Å.
HYDROGEN BONDS are a very strong dipole dipole interaction that arises when you have H
bonded to N, O or F, attracted to N, O or F. We are talking energies on the order of 20 kJ/mol for Hbonds at 5.00 Å (see p R9.3-9.4).
d. London forces are the only forces of coulombic (electrostatic) attraction between two oppositely
charged ends of an induced dipole in a molecule which is on average nonpolar. All molecules can do
this. The potential energy of attraction varies as 1/r6. That is, energy holding them together is
inversely proportional to the 6th power of the distance apart, and directly proportional to the product
of the charges, and both of the charges will be less than 1. London forces are very weak, say 1
kJ/mol at 5 Å. In general it is easier to induce a dipole in a larger atom than a smaller one.
IN GENERAL: Compounds which are solids at room temp are held together by stronger
intermolecular forces than compounds which are liquids. (compare sodium chloride and water)
Compounds which are gases at RT are held together by weaker intermolecular forces still. (Compare
water and O2 ).
9.
See answer above: "IN GENERAL: Compounds which are solids at room temp are held together by
stronger intermolecular forces than compounds which are liquids. (compare sodium chloride and water)
Compounds which are gases at RT are held together by weaker intermolecular forces still. (Compare water and
O2 )." Compound Y is a liquid at RT, it must be the more polar one, because dipole-dipole
interactions are stronger than London forces, which are the only intermolecular forces available to a
nonpolar compound.
10. In general it will take more energy to hoist a large molecule into the gas phase than a small one.
Large molecules have more mass, and, being larger, have more polar areas or more easily form
induced dipoles. Molecule Y will have the higher boiling pt.
12.
As you would predict from your own experience: Squash it set it out in the snow and you get
a solid.
Release the pressure & heat it up, you get a gas.
P T  solid
P  T  gas
19. Molar volume has units of L/mole. Flip density over and convert to L and moles:
(1mL /10.59 g Ag)(1L/1000 mL)(107.868 g Ag/mole Ag) = 0.01019 L/mole Ag
To get particles of Ag/mm3, flip the molar volume term over:
(1moleAg/0.01019 L)(6.022 x 1023particles Ag/mole ag)(1 L/1000 mL)(1 mL/1cm3)([1 cm/10 mm]3) =
5.910 x 1019 Ag atoms/mole.
AW
Density
Molar Volume
(g/mol)
(g/mL)
(L/mol)
Ag
107.868
10.59
1.019E-02
5.910E+19
Bi
208.98
9.80
2.13E-02
2.82E+19
Ne
20.18
0.00090
2.2E+01
2.7E+16
particles/mm3
21. [1 mole/22.06 L)*(70.90g/mol)*(1L/1000 mL) = g/mL, which is density.
(1 mol/22.06L)*(6.02 x 1023 particles/mol) = particles/L
NOTICE HOW FEW PARTICLES/L Cl2 HAS (A GAS) COMPARED TO Br2 (A LIQUID) AND I2
(A SOLID)
AW
MW
Mol. Vol
Density
moles/L
particles/L
(g/mol)
(g/mol)
(L/mol)
g/mL
Cl2
35.45
70.90
22.06
0.003214
0.05
2.73E+22
Br2
79.90
159.80
0.0512
3.12
19.53
1.18E+25
I2
126.90
253.80
0.0515
4.93
19.42
1.17E+25
23. Na2SO4 --> 2Na+ + SO4=
(36.4 gNa2SO4/463 mL)*(1 mol Na2SO4/142.042g)*(2 Na+/Na2SO4)*(1000 mL/L) = moles Na+/L
MW
Na2SO4
(g/mol)
142.042
g Na2SO4
36.4
Vol. Soln
(mL)
463
[Na+]
(mol/L)
1.11
24. 1 mL * (1L/1000 mL) * (1.11 moles Na+/L)(1 SO4=/2 Na+) (6.02 x 1023 particles/mol) = #
SO4=/mL
3.33E+20SO4= ions in a mL
24. 1.00 M dextrose means 1.00 mole of dextrose in 1.00 L of solution (NOT 1.00 L of water!!!!).
The author is sloppily implying that you can calculate mole fractions exactly if
only molar concentration is known. It can't, precisely. However, in reasonably dilute solutions,
the mole fraction can be approximated well.
You have 1 mole of dextrose and 1 L of solution. You must make the approximation that the L of
solution is MOSTLY water. (The actual volume of water is slightly less than a liter. The dextrose
takes up some room. However, one works with what one has...)
So, you have roughly 1 L of water. How many moles? Use the density of water & convert to
moles:
(1 g H2O/mL)(1000 mL/L)(1 mole H2O/18.0 g) = 55.5 moles H2O/L
Now mole fraction is a piece of cake:
X Dextrose: 1mole/(1 mole + 55.5 moles)
X Water:
55.5 moles/(1 mole + 55.5 moles)
Mole Fraction dextrose:
0.0177
Mole Fraction water:
0.982

1.0000
25. SAMPLE CALCULATION:
a. 10 mL MeOH*(0.7914 g/mL)(1 mol MeOH/32.0 g) = moles MeOH
10 mL EtOH*(0.7893 g/mL)(1 mol EtOH/46.0 g) = moles EtOH
X MeOH = moles MeOH/(Moles MeOH + moles EtOH)
X EtOH = moles EtOH/(Moles MeOH + moles EtOH)
a.
MeOH
EtOH
H2O
Density
g/mL
0.7914
0.7893
0.9982
Volume
mL
10.0
10.0
0.0
b.
MeOH
EtOH
H2O
10.0
0.00
10.0
0.247
0.000
0.555

0.308
0.000
0.692
1.000
c.
MeOH
EtOH
H2O
0.0
10.00
10.0
0.000
0.172
0.555

0.000
0.236
0.764
1.000
d.
MeOH
EtOH
H2O
10.0
10.0
10.0
0.247
0.172
0.555

0.254
0.176
0.570
1.000
e.
MeOH
EtOH
H2O
72.0
1050
413.0
1.781
0.018
22.903

0.072
0.001
0.927
1.000
MeOH
EtOH
H2O
MW
(g/mol)
32.0
46.0
18.0
moles
0.247
0.172
0.000


0.590
0.410
0.000
1.000
27. a. 2.00 g Li2SO4*/50.00 mL sol'n) *(1 mol Li2SO4/109.948 g)(1000 mL/L) = molarity of Li2SO4
MW
g
(g/mol)
Li2SO4
109.948
2.00
ml sol'n
moles
M
Li2SO4
50.00
0.0182
0.364
b. Once again, the author is rather sloppily implying that you can calculate mole fractions exactly if
only molar concentration is known. It can't, precisely. However, in reasonably dilute solutions, the
mole fraction can be approximated.
You have 0.0182 mole of Li2SO4 and 50.0 mL of solution. You must make the approximation that the
50 mL of solution is MOSTLY water. (The actual volume of water is slightly less. The Li2SO4 takes
up some room. However, one works with what one has...)
So, you have roughly 50 mL of water. How many moles? Use the density of water & convert to
moles:
(1 g H2O/mL)(50.0 mL)(1 mole H2O/18.0 g) = moles H2O
You have 0.0182 moles of SO4= & 2x0.0182 moles of Li+
XH2O = moles H2O /(moles H2O + moles Li+ + Moles SO4=)
moles
X
H2O
2.778
0.981
Li+
0.0364
0.013
SO4=
0.0182
0.006

1.000
28. You have 0.0182 moles of SO4= & 2x0.0182 moles of Li+. You also now have approximately 1.00
L of water, which is 55.5 moles of water:
(1 g H2O/mL)(1000 mL/L)(1 mole H2O/18.0 g) = 55.5 moles H2O/L
XH2O = moles H2O /(moles H2O + moles Li+ + Moles SO4=)
moles
X
H2O
55.556
0.99902
Li+
0.0364
6.542E-04
SO4=
0.0182
3.271E-04

1.000
29. When in doubt convert to moles:
Solution A: 50.0 mL(1L/1000 mL)(0.500 moles Li2SO4= ) /L) = 0.0250 moles Li2SO4
Solution B: 100.0 mL(1L/1000 mL)(0.250 moles Li2SO4= ) /L) = 0.0250 moles Li2SO4
Total moles Li2SO4= after mixing: add ‘em up: 0.0250 moles + 0.0250 moles = 0.0500 moles Li2SO4.
Total volume after mixing: add ‘em up: 50.0 mL + 100.0 mL = 150.0 mL(1 L/1000 mL) = 0.150 L
[Li2SO4] = 0.0500 moles Li2SO4=/0.150 L = 0.333 M
30 a. The anion is the same for all salts. Thus the salt with the smallest cation will have the highest
lattice energy, as the small size means that the + & - charges are closest together:
THUS: LiCl > NaCl > KCl > RbCl
b.
The cation is the same for all salts. Thus the salt with the smallest anion will have the
highest lattice energy, as the small size means that the + & - charges are closest together:
THUS: AgF > AgCl > AgBr > AgI
c. Na+ …
Ca2+ …
Mg2+ …
ClS=
S=
The salt with the ions of smallest charge (NaCl) will have the lowest lattice energy:
THUS X > Y > NaCL
Of CaS & MgS, the anion is the same, the valence (charge) on the cation is the same
So the salt with the smaller cation (Mg2+ ) will have the highest lattice energy:
THUS: MgS > CaS > NaCl
30.
a.
Ca2+ …CO3= - IONIC
b.
Pt – a metal… nuclei in a sea of electrons.
c.
N2 Nonpolar LONDON FORCES
d.
H2O H BONDS
31. H-bonds!
LDPE has only London forces holding its strands together, since it is a non-polar molecule.
33a.
35. Well, Ag is a metal, and is the only solid at room temp. on the list, (Highest mp & bp).
So :
Ag > all the rest
Water is the only liquid at RT and not coincidentally the only one of the nonmetals to be able to H
bond.
So: Ag > H2O > the gases.
Of the gases, He is the smallest, its London Forces will be weakest, so it will be the lowest MP, & BP:
So: Ag > H2O > …> He.
Neon is smaller than CO2,
So: Ag > H2O >CO2 >Ne > He.
36. CH3OH is the only one on the list which can H-bond; CF4 and CH4 are both non-polar and
can only use London forces.
So: CH3OH is the highest boiling.
Between CF4 and CH4, the larger molecule, CF4 , will have the stronger London forces and be higher
boiling than CH4.
So:
CH3OH >
CF4 > CH4