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Tardigrade
Question
Chemistry
The K sp for bismuth sulphide ( Bi 2 S 3) is 1.08 × 10-73. The solubility of Bi 2 S 3 in mol L -1 at 298 K is
Q. The
K
s
p
for bismuth sulphide
(
B
i
2
S
3
)
is
1.08
×
1
0
−
73
. The solubility of
B
i
2
S
3
in
m
o
l
L
−
1
at
298
K
is
990
160
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Equilibrium
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A
1.0
×
1
0
−
15
67%
B
2.7
×
1
0
−
12
0%
C
3.2
×
1
0
−
10
33%
D
4.2
×
1
0
−
8
0%
Solution:
k
s
p
=
(
2
s
)
2
(
3
s
)
3
=
4
s
2
×
27
(
s
)
3
=
108
(
s
)
5
(
s
)
5
=
108
1.08
×
1
0
−
73
⇒
s
=
1
0
−
15