Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
Ksp of CaSO4⋅5H2O is 9 × 10-6, find the volume for 1 g of CaSO4 (M.wt. = 136).
Q.
K
s
p
of
C
a
S
O
4
⋅
5
H
2
O
is
9
×
1
0
−
6
, find the volume for
1
g
of
C
a
S
O
4
(M.wt. = 136).
3498
212
AIIMS
AIIMS 2011
Equilibrium
Report Error
A
2.45 litre
59%
B
5.1 litre
8%
C
4.52 litre
23%
D
3.2 litre
10%
Solution:
C
a
S
O
4
(
s
)
⇌
S
C
a
2
+
(
a
q
)
+
S
S
O
4
2
−
(
a
q
)
K
s
p
=
S
2
=
9
×
1
0
−
6
S
=
3
×
1
0
−
3
m
o
l
L
−
1
Solubility in
g
l
i
t
r
e
−
1
molecular mass
×
S
=
136
×
3
×
1
0
−
3
=
408
×
1
0
−
3
g
L
−
1
408
×
1
0
−
3
g
of
C
a
S
O
4
present in 1 litre
1
g
of
C
a
S
O
4
present is
408
×
1
0
−
3
1
=
2.45
litre