Solutions to exercises
29
or
y
x
x
1
1
1
− = +
−
which gives
y
x
x
1
2
= +
2. For x = 0.08, we have y = 0.148. The formulas for the solid solution are
(ZrO 2 ) 0.92 (Y 2 O 3 ) 0.08
and
(ZrO 2 ) 0.852 (YO 1.5 ) 0.148
3. For a formula unit, the number of cationic sites in each case is
for (ZrO 2 ) 0.92 (Y 2 O 3 ) 0.08
n Zr = 0.92 + 2 # 0.08
or
n Zr = 1.08
for (ZrO 2 ) 0.852 (YO 1.5 ) 0.148
n Zr = 0.852 + 0.148
or
n Zr = 1
4. a. Expression of the formula (ZrO 2 ) 1−x (Y 2 O 3 ) x in the form Zr α Y β O γ V δ with
α, β, γ, and δ expressed as a function of x.
We first transform the formula into the form Zr 1−x Y 2x O 2(1−x)+3x V δ
By identification, we have α = 1 − x, β = 2x, and γ = 2 + x
For δ, the number of anionic sites is twice that of cationic sites, from
which we deduce
δ + 2(1− x) + 3x = 2 (1 + x)
or
δ = x
Thus, (ZrO 2 ) 1−x (Y 2 O 3 ) x may be written in the form
Zr 1–x Y 2x O 2+x V x
The numerical application to (ZrO 2 ) 1−x (Y 2 O 3 ) x for x = 0.08 gives
Zr 0.92 Y 0.16 O 2.08 V 0.08
b. Expression of the formula (ZrO 2 ) 1–y (YO 1.5 ) y in the form Zr α Y β O γ V δ
with α, β, γ and δ expressed as a function of x
We transform the formula into the form Zr 1−y Y y O 2(1−y)+ 2
3 y V δ .
By identification, we have α = 1 − y, β = y, and γ = 2 − 2
1
y.