Chapter 9: Liquids and Solids
10. Refer to Section 9.1 and Example 9.2.
P = 681 mm Hg x
P
ln 2
 P1
1 atm
= 0.896 atm
760 mm Hg
 + ∆ H vap
 =
R

1
1
 −
 T1 T2



4
1
 0.896 atm  4.07 x 10 J/mol.  1

−
ln
=

8.31 J/mol. ⋅ K  373 K T2
 1.00 atm 

1
- 0.110 = (4.90 x 10 3 K) 0.00268 K -1 −
T2

- 2.25 x 10 -5 K -1 = 0.00268 K -1 −
1
= 0.00270
T2
T2 = 370 K = 97°C
1
T2






18. Refer to Section 9.2.
a.
112 atm
C
solid
liqu id
~ 40 atm
A = triple point
B = normal boiling point
C = critical point
M = normal melting point
Pressure
M
1 atm
B
vapor
46 mm Hg
(Note - Not drawn to scale)
A
-78
-33
40
132
T (°C)
b. As shown in the phase diagram above, the pressure will be about 40 atm. The value one
gets will depend on the curve draw for the vapor-liquid boundary, but must be between 1
atm (normal boiling point) and 112 atm (critical point).
32. Refer to Section 9.3 and Example 9.4.
a. LiCl or CCl4. LiCl is an ionic compound, which typically have high melting and boiling
points, while CCl4 has only dispersion forces, and thus, the lower boiling point.
b. CH3OH or CH3F. Both of these compounds have both dispersion and dipole forces.
CH3OH, however, also has hydrogen bonding, and has a higher boiling point.
c. H2O or SO2. Both of these compounds have both dispersion and dipole forces. H2O,
however, also has hydrogen bonding, and has a higher boiling point.
d. N2 or Cl2. Both of these molecules have only dispersion forces. N2, however, has fewer
electrons and thus, weaker dispersion forces and a lower boiling point.
50. Refer to Section 9.5, Figures 9.17 and 9.18, and Example 9.8.
K+ : r = 0.133 nm
I- : r = 0.216 nm
a. one side of a cube
K+ and I- ions touch along an edge of a cell
s = 0.216 nm + 2(0.133 nm) + 0.216 nm = 0.698 nm
b. face diagonal
d =s 2
d = 0.698(1.41)
d= 0.987 nm