4.10 The concentration of an element in an alloy, in atom percent, may be computed using
Equation (4.5). With this problem, it first becomes necessary to compute the number of
moles of both Cu and Zn, for which Equation (4.4) is employed. Thus, the number of moles
of Cu is just
nm
=
Cu
m'
Cu
33 g
= 0.519 mol
63.55 g / mol
=
A Cu
Likewise, for Zn
nm
=
Zn
47 g
= 0.719 mol
65.39 g / mol
Now, use of Equation (4.5) yields
'
C Cu
=
=
nm
Cu
 nm
Cu
Zn
nm
x 100
0.519 mol
x 100 = 41.9 at%
0.519 mol  0.719 mol
Also, 4.10 The concentration of an element in an alloy, in atom percent, may be computed
using Equation (4.5). With this problem, it first becomes necessary to compute the number
of moles of both Cu and Zn, for which Equation (4.4) is employed. Thus, the number of
moles of Cu is just
nm
=
Cu
m'
Cu
A Cu
=
33 g
= 0.519 mol
63.55 g / mol
Likewise, for Zn
nm
=
Zn
Now, use of Equation (4.5) yields
47 g
= 0.719 mol
65.39 g / mol
'
C Zn
=
4.14
0.719 mol
x 100 = 58.1 at%
0.519 mol  0.719 mol
This problem calls for a determination of the number of atoms per cubic meter for
aluminum. In order to solve this problem, one must employ Equation (4.2),
N =
NA  Al
A Al
3
The density of Al (from the table inside of the front cover) is 2.71 g/cm , while its atomic
weight is 26.98 g/mol. Thus,
N =
6.023

x 10 23 atoms/ mol 2.71 g / cm 3

26 .98 g / mol
= 6.05 x 10
22
3
28
3
atoms/cm = 6.05 x 10 atoms/m
3
4.15 In order to compute the concentration in kg/m of Si in a 0.25 wt% Si-99.75 wt% Fe alloy
we must employ Equation (4.9) as
CSi
''
CSi
=
CSi
Si

CFe
x 103
Fe
3
From inside the front cover, densities for silicon and iron are 2.33 and 7.87 g/cm ,
respectively; and, therefore
0.25
''
CSi
=
0.25
2.33 g / cm3

99 .75
x 103
7.87 g / cm3
= 19.6 kg/m
3
4.18 This problem asks us to determine the number of molybdenum atoms per cubic centimeter
for a 16.4 wt% Mo-83.6 wt% W solid solution.
To solve this problem, employment of
Equation (4.17) is necessary, using the following values:
C =C
= 16.4 wt%
1
Mo
3
 =
= 10.22 g/cm
1
Mo
3
 =  = 19.3 g/cm
2
W
A =A
= 95.94 g/mol
1
Mo
Thus
NMo =
NA CMo
CMo A Mo
A Mo



Mo
=
6.023
W
100
 CMo


x 10 23 atoms / mol (16.4)
(16.4)(95.94 g / mol)
(10.22 g / cm 3 )

= 1.73 x 10
95.94 g / mol
19.3 g / cm 3
22
atoms/cm
100
 16.4 
3
4.22 This problem asks us to determine the weight percent of Ge that must be added to Si such
21
that the resultant alloy will contain 2.43 x10 Ge atoms per cubic centimeter. To solve this
problem, employment of Equation (4.18) is necessary, using the following values:
21
3
N =N
= 2.43 x 10 atoms/cm
1
Ge
3
 =
= 5.32 g/cm
1
Ge
3
 =  = 2.33 g/cm
2
Si
A =A
= 72.59 g/mol
1
Ge
A = A = 28.09 g/mol
2
Si
Thus
C Ge =
1
100
N A Si
N GeA Ge

Si
 Ge
100
=
6.023 x10 atoms/ mol(2.33 g / cm3 )  2.33
2.43 x10 21 atoms / cm3 (72.59 g / mol) 5.32
23
1
g/ cm 3 


g/ cm 3 
------------------------------------------------------------------------------= 11.7 wt%
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