The conductivity of saturated solution of AgCl is found to $1 .86 \times 10^{- 6} {ohm}^{- 1} {cm}^{- 1}$ and that of water is $6 \times 10^{- 8} {ohm}^{- 1} {cm}^{- 1} . If λ^{0} AgCl is 138$ ${ohm}^{- 1} {cm}^{2} {eq}^{- 1}$ the solubility product of AgCl is

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1.69 × 10^–10

2 × 10^–10

1.3 × 10^–5

2.7 × 10^–10

answer is B.

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Detailed Solution

Conductivity of AgCl = Conductivity of solution - Conductivity of water

= 1.8 $\times$ 10^-6 ohm^-1cm^-1

$λ_{M}^{0} = \frac{κ \times 1000}{M}$

$M = \frac{1.8 \times 10^{- 6} \times 1000}{138} = 0.013 \times 10^{- 3} = 1.3 \times 10^{- 5}$

K_sp = ${(1.3 \times 10^{- 5})}^{2} = 1.69 \times 10^{- 10}$

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