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Solutions to Chapter 11 Practice Problems (Watts).
Note: these are my solutions. I have worked them out and typed them from scratch,
which takes me a lot of time and effort. They have not been copied from the textbook
publisher’s materials.
Problem 9
The intermolecular forces between H 2 O molecules are much stronger than those between
CH 4 molecules.
Problem 18
Vaporization of CS 2 ( l ): CS 2 ( l ) → CS 2 (g)
We need the enthalpy of vaporization: ΔHvap = 27.4 kJ/mol (see Table 11.1, page 435)Hvap = 27.4 kJ/mol (see Table 11.1, page 435)
To find the heat need to vaporize 275 μL, we must find out how many moles there are. L, we must find out how many moles there are.
We are given the volume and density. Find mass first.
M = dV = (1.26 g/mL) 275 x 10-3^ mL = 0.3465 g
Number of moles: 4.^55010 mol
15 g mol
3465 g 3 1
MM
m n
Heat required: (4.550 x 10
- mol) (27.4 kJ/mol) = 0.125 kJ
Problem 22
The equation given tells us the ΔHvap = 27.4 kJ/mol (see Table 11.1, page 435)H for combustion of one mole of pentane (C 5 H 12 ) is -
3.51 x 10
3 kJ, i.e. combustion of one mole of pentane produces 3.51 x 10
3 kJ of heat. We
need to find how much heat will be produced from the combustion of 1.25 g of pentane.
Number of moles of pentane in 1.25 g: 0.^0173 mol
17 g mol
25 g (^) MM 1
m n
Heat required to vaporize 165 g of hexane (C 6 H 14 ):
(0.0173 mol) 3.51 x 10
3 kJ/mol = 60.7 kJ
How many moles of hexane is 165 g? 1.^91 mol
- 20 g mol
165 g n (^) 1
Hence, 60.7 kJ can vaporize 1.91 mol of hexane. Therefore, the heat required to vaporize
1 mol of hexane is 60.7 kJ / 1.91 mol = 31.8 kJ/mol: ΔHvap = 27.4 kJ/mol (see Table 11.1, page 435)Hvapn = 31.8 kJ/mol.
Problem 28
This involves a gas law calculation on silver vapor. We are given the volume (486 mL =
0.486 L), temperature (
o C = 1633.15 K), and pressure (1.00 mmHg = 1.32 x 10
atm).
We use the ideal gas equation to find the number of moles of Ag atoms. Then we use
Avogadro’s number to find the number of Ag atoms.
- 78 10 mol
- 0821 Latmmol K 1633. 15 K
( 1. 32 10 atm)( 0. 486 L ) 6 1 1
3
RT
PV n
Number of atoms: N = n NA = (4.68 x 10-6^ mol)(6.022 x 10^23 mol-1) = 2.82 x 10^18
Problem 30
We can use the relationship between pressure, density, molar mass, and temperature of a
gas. As we recall from Chapter 5:
P
dRT MM (^) so MM
dRT P
Density of acetone vapor: d = 0.876 g/L; temperature: T = 32
o C = 305.15 K; molar mass
= 58.09 g mol
378 atm 287 mmHg
09 g mol
876 g/L 0. 0821 Latmmol K 305. 15 K 1
1 1
MM
dRT P
Problem 38
We first have to calculate the heat of combustion of 3.12 L of CH 4 (g) (at 25 oC and 753
mmHg). To do this, we must first calculate the number of moles of CH 4 , then use the
given ΔHvap = 27.4 kJ/mol (see Table 11.1, page 435)H for combustion of 1 mol (-890.3 kJ).
T = 25 oC = 298.15 K. P = 753 mmHg = 0.991 atm. V = 3.12 L.
126 mol
0821 Latmmol K 298. 15 K
991 atm 3. 12 L 1 1 RT
PV n
Heat of combustion of this much CH 4 :
ΔHvap = 27.4 kJ/mol (see Table 11.1, page 435)H = 0.126 mol (-890.3 kJ/mol) = -112 kJ
The final issue is what mass of ice can be melted by 112 kJ. The ΔHvap = 27.4 kJ/mol (see Table 11.1, page 435)Hfusion of ice is 6.01 kJ/
mol. So how many moles of ice could 112 kJ melt?
for Cl 2 and CCl 4 , Cl 2 will have the lower melting point (smaller dispersion forces than
CCl 4 ). Overall ranking in order of increasing melting point:
Cl 2 < CCl 4 < CsCl < MgCl 2
Problem 60
Most likely ethylene glycol (EG) has the higher surface tension. It would probably has
stronger intermolecular forces than isopropyl alcohol (IA): their molar masses are not
much different, but EG has 2 OH groups, while IA has only one.
I looked up the data to check. The surface tensions (in units of dyne/cm) are:
Ethylene glycol 47.7 Isopropyl alcohol 21.
Problem 68
The face-centered cubic unit cell is shown on page 472 (Figure 11.47). It contains 4 K+
ions and 4 Cl- ions:
Green ions: 8 x 1/8 (from the 8 vertices (corners)) + 6 x ½ (from the 6 faces)
Purple ions: 1 in the middle + 4 x ¼ (top) + 4 x ¼ (bottom) + 4 x ¼ (middle)
Mass of the unit cell of KCl:
4 x (39.0983 + 35.4527) u = 298.2040 u
Length of one side of the unit cell:
2 x 314.54 pm = 629.08 pm = 6.2908 x 10-10^ m
Volume of one unit cell: (cube the length of one side)
2.4895 x 10-28^ m^3 = 2.4895 x 10-22^ cm^3
Density of KCl (in units of u cm
- ): (divide mass of unit cell by volume)
24 3 22 3 1.^197810 ucm
4895 10 cm
2040 u
V
m d
Since the measured density is 1.9893 g cm
24 u = 1.9893 g. Hence, based on
the measured density and internuclear distance, Avogadro’s number is 1.1978 x 10^24 /
1.9893 = 6.0212 x 10
23 .
Problem 70
Based on the figure, there are 4 Ca
2+ ions per unit cell (8 x 1/8 from the cornbers; 6 x ½
from the faces) and 8 F
- ions per unit cell. Hence, the formula CaF 2 is consistent with the
unit cell.
Coordination numbers: 8 for Ca2+; 4 for F-. We wouldn’t expect them to be the same
because of the stoichometry. For a compound like NaCl, we would expect the same
coordination numbers.
