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Byju's Answer
Standard XII
Mathematics
Properties of Conjugate of a Complex Number
K sp of Zr 3 ...
Question
K
s
p
of
Z
r
3
(
P
O
4
)
4
in terms of solubility (s) is:
A
108
s
7
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B
4
s
3
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C
6912
s
7
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D
27
s
4
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Solution
The correct option is
C
6912
s
7
Z
r
3
(
P
O
4
)
4
⇌
3
Z
r
4
+
+
4
P
O
4
3
−
3
s
4
s
K
s
p
=
(
Z
r
4
+
)
3
(
P
O
3
−
4
)
4
K
s
p
=
(
3
s
)
3
(
4
s
)
4
K
s
p
=
27
s
3
×
256
s
4
K
s
p
=
6912
s
7
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