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Question

The specific conductance of a saturated solution of AgCl is “κΩ1cm1. The limiting ionic conductance of Ag+ and Cl ions are x and y respectively. Then, the solubility product of AgCl is:

A
100κ(x+y)
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B
(1000κ)2(x+y)2
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C
(1000×14.3×κ)(x+y)
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D
(103×143.5×κ)2(x+y)2
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Solution

The correct option is B (1000κ)2(x+y)2
Point to remember:
Solublity product:
ABSA+S+BS
Ksp=S2
λ0m=1000κ/C
C=S
given data:
λ0Ag+=x
λ0Cl=y
so λ0AgCl=x+y
On substituting we get
λ0AgCl=1000κ/S
S=1000κx+y
So Ksp=(1000κx+y)2

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