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Tardigrade
Question
Chemistry
The solubility product of AgCl is 1.8 × 10-10, the minimum volume ( in L ) of water required to dissolve 1.9 mg of AgCl is approximately.( molecular weight of AgCl =143.5 )
Q. The solubility product of
A
g
Cl
is
1.8
×
1
0
−
10
, the minimum volume
(
in
L
)
of water required to dissolve
1.9
m
g
of
A
g
Cl
is approximately.( molecular weight of
A
g
Cl
=
143.5
)
2913
210
Equilibrium
Report Error
A
10
B
2
C
1
D
20
Solution:
Solubility
=
1.8
×
1
0
−
5
M
=
1.34
×
1
0
−
5
M
=
1.34
×
1
0
−
5
×
143.5
g
m
/
l
i
t
re
=
192.29
×
1
0
−
5
g
m
/
l
i
t
re
=
1.92
m
g
/
l
i
t
re
so required volume
=
1
L