MSE-227
Answers
Homework Set #3
Chapter 4
4.2
(a) 100.02 g/mol
(b) 100.11 g/mol
(c) 226.32 g/mol
(d) 192.16 g/mol
4.6
(a) M n  47,720 g / mol
(b) M w  53,720 g / mol
(c) m 
(d)
M n 47,720 g / mol

 100.04 g / mol
nn
477
Chapter 5:
Chapter 6:
Dislocations
Edge Dislocation
Screw Dislocation
It is harder to move a dislocation through NaCl because the atoms have to be displaced
farther to come back into registry with like neighbors (in order to maintain charge
neutrality). The energy required goes up as the square of the distance moved.
Composition
If the alloy is initially at 90% Silver (Ag) and 10% Copper (Cu), then, for 100 moles of
material, there are 90 mol of Ag and 10 mol of Cu. If we increase the amount of Cu to
make the alloy 12 at% Cu, then we need to solve for the following:
xCu
 0.12
Solving for x gives x = 12.27 mol Cu
90 Ag  xCu
Cu has a molecular weight, MW = 63.55 g/mol, and Ag = 107.87 g/mol.
The percentage increase in the number of mol of Cu is:
(12.27-10)/10 = 22.73% increase. This will also be the increase in the weight of Cu in
the alloy.
We need to solve for the initial weight of Cu in an alloy that has a total weight of 50g:
10molCu  63.55 g / mol
 100  6.14wt %Cu
90molAg *107.87 g / mol  10molCu * 63.55 g / mol
An alloy that has 100 grams total, then there will be 6.14 g Cu and 93.86 g Ag.
Therefore, an alloy of 50 grams will have half these amounts: 3.07 g. Cu and 46.93 g Ag.
Our new alloy will have 22.73% more Cu, for a total of 1.2273 * 3.07 = 3.7678 g Cu.
The amount we need to add, then, is 3.7678 g – 3.07 g = 0.6978 g Cu.
Note: those of you that check the answer, using the numbers here, will find it to be 11.8
at% Cu, not 12, if we add 0.6978 g Cu. This is a rounding error; doing the problem with
more significant figures gives the correct number, 12.00 at% Cu.