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Tardigrade
Question
Chemistry
The solubility product of AgCl is 1.5625 × 10-10 at 25 °C. Its solubility in grams per litre will be
Q. The solubility product of
A
g
Cl
is
1.5625
×
1
0
−
10
at
25
∘
C
. Its solubility in grams per litre will be
5890
194
Equilibrium
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A
143.5
7%
B
108
8%
C
1.57
×
1
0
−
8
31%
D
1.79
×
1
0
−
3
54%
Solution:
A
g
Cl
⇌
s
A
g
+
+
s
C
l
−
s
2
=
1.5625
×
1
0
−
10
s
=
1.25
×
1
0
−
5
m
o
l
L
−
1
Solubility in
g
L
−
1
=
Molar mass
×
s
=
143.5
×
1.25
×
1
0
−
5
=
1.79
×
1
0
−
3
g
L
−
1