Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
The solubility of CaF2 in pure water is 2.3 × 10- 6 mol dm-3 . Its solubility product will be :
Q. The solubility of
C
a
F
2
in pure water is
2.3
×
1
0
−
6
m
o
l
d
m
−
3
. Its solubility product will be :
2076
186
UPSEE
UPSEE 2006
Report Error
A
4.8
×
1
0
−
18
B
48.66
×
1
0
−
18
C
4.9
×
1
0
−
11
D
48.66
×
1
0
−
15
Solution:
C
a
F
2
s
Ca
X
2
+
+
2s
2
F
X
−
K
SP
=
s
(
2
s
)
2
=
4
s
3
K
SP
=
4
(
2.3
×
1
0
−
6
)
3
48.668
×
1
0
−
18
(
m
o
l
d
m
−
3
)
3