The specific conductance of a saturated solution of silver chloride is ${KΩ}^{- 1} {cm}^{- 1}$ .The limiting ionic conductance of ${Ag}^{+} & {Cl}^{-}$ ions are x and y, respectively. The solubility of silver chloride in $g {litre}^{- 1}$ is : (molar mass of AgCl =143.5 )

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$\frac{K \times 1000 \times 143.5}{x + y}$

$\frac{K \times 1000}{x - y}$

$\frac{x + y}{K} \times \frac{1000}{143.5}$

$\frac{K}{x - y} \times 143.5$

answer is C.

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Detailed Solution

Specific conductance = k

$λ_{AgCl}^{\infty} = λ_{{Ag}^{+}}^{\infty} + λ_{{cl}^{-}}^{\infty} = x + y$

$λ_{AgCl}^{\infty} = \frac{K \times 1000}{s (moles / litre)} \Rightarrow S = \frac{K \times 1000}{X + y} \times 143.5 g / litre$

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