1. Callister 4.34
(a) This part of problem asks that we compute the number of grains per square inch for an ASTM grain
size of 8 at a magnification of 100. All we need do is solve for the parameter N in Equation 4.16,
inasmuch as n = 8. Thus
N  2n1


= 281 = 128 grains/in.2
(b) Now it is necessary to compute the value of N for no magnification. In order to solve this problem it is
necessary to use Equation 4.17:
 M 2
N M    2 n1
100 

where NM = the number of grains per square inch at magnification M, and n is the ASTM grain size
number. Without any magnification, M in the above equation is 1, and therefore,
 1 2
N1    281  128
100 

And, solving for N1, N1 = 1,280,000 grains/in.2.
2. Callister 14.2
(a) For poly(vinyl chloride), each repeat unit consists of two carbons, three hydrogens, and one chlorine
(Table 14.3). If AC, AH and ACl represent the atomic weights of carbon, hydrogen, and chlorine,
respectively, then
m = 2(AC) + 3(AH) + (ACl)
= (2)(12.01 g/mol) + (3)(1.008 g/mol) + 35.45 g/mol = 62.49 g/mol
(b) For poly(ethylene terephthalate), from Table 14.3, each repeat unit has ten carbons, eight
hydrogens, and four oxygens. Thus,
m = 10(AC) + 8(AH) + 4(AO)
= (10)(12.01 g/mol) + (8)(1.008 g/mol) + (4)(16.00 g/mol) = 192.16 g/mol
(c) For polycarbonate, from Table 14.3, each repeat unit has sixteen carbons, fourteen hydrogens,
and three oxygens. Thus,
m = 16(AC) + 14(AH) + 3(AO)
= (16)(12.01 g/mol) + (14)(1.008 g/mol) + (3)(16.00 g/mol)
= 254.27 g/mol
(d) For polydimethylsiloxane, from Table 14.5, each repeat unit has two carbons, six hydrogens,
one silicon and one oxygen. Thus,
m = 2(AC) + 6(AH) + (ASi) + (AO)
= (2)(12.01 g/mol) + (6)(1.008 g/mol) + (28.09 g/mol) + (16.00 g/mol) = 74.16 g/mol
3. Callister 14.11
We are asked to sketch portions of a linear polystyrene molecule for different configurations (using twodimensional schematic sketches).
(a) Syndiotactic polystyrene
(b) Atactic polystyrene
(c) Isotactic polystyrene
4. From summer 2010 midterm
5. Callister 12.13
We are asked to calculate the theoretical density of FeO. This density may be computed using Equation (12.1) as
nAFe + AO 
VC N A
 
Since the crystal structure is rock salt, n' = 4 formula units per unit cell. Using the ionic radii for Fe2+ and O2- from

Table 12.3, the unit cell volume is computed as follows:

VC  a 3  2rFe 2+ + 2rO 2-
= 0.0817

Thus,

3
3
 2 (0.077 nm)  2 (0.140 nm)
nm3
cm 3
= 8.17  10-23
unit cell
unit cell

 

(4 formula units/unit cell)(55.85 g/mol + 16.00 g/mol)
8.17  10-23 cm3/unit cell6.022  1023 formula units/mol
= 5.84 g/cm3
6. Callister 12.24
We are asked in this problem to compute the atomic packing factor for the CsCl crystal structure. This
requires that we take the ratio of the sphere volume within the unit cell and the total unit cell volume. From
Figure 12.3 there is the equivalence of one Cs and one Cl ion per unit cell; the ionic radii of these two ions
are 0.170 nm and 0.181 nm, respectively (Table 12.3). Thus, the sphere volume, VS, is just
VS 


2r + + 2r Cs
Cl
3

2(0.170 nm) + 2(0.181 nm)
3
= 0.405 nm
Since VC = a3
VC = (0.405 nm)3 = 0.0664 nm3
And, finally the atomic packing factor is just
V
0.0454 nm3
APF  S 
 0.684
VC
0.0664 nm3

0.0454 nm3
For CsCl the unit cell edge length, a, in terms of the atomic radii is just
a 


4
() (0.170 nm)3 + (0.181 nm)3
3
7. From summer 2010 midterm
8. Callister 13.D1
(a) Important characteristics that are required of a ceramic material to be used for kitchen cookware are:
(1) it must have a high resistance to thermal shock (Section 19.5) in order to withstand relatively rapid
changes in temperature; (2) it must have a relatively high thermal conductivity; 3) it must be relatively
strong and tough in order to endure normal kitchen use; and 4) it must be nontoxic.
(b) Possible materials worth considering are a common soda-lime glass, a borosilicate (Pyrex) glass, and a
glass ceramic. These materials and some of their characteristics are discussed in this chapter. Using
Equation 17.9 a comparison of the resistance to thermal shock may be made. The student will need to
obtain cost information.
(c) It is left to the student to make this determination and justify the decision.