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The specific conductance of a saturated solution of AgCl in water is 1.826 x 10^–6ohm^–1cm^–1 at 25°C. Calculate its solubility in water at 25°C. [Given Λ^∞_m (Ag⁺) = 61.92 ohm^–1 cm² mol^–1and Λ^∞_m (CI^– ) = 76.34 ohm^–1 cm² mol^–1]

Solution

Λ^∞_m (AgCl) = λ^∞_m_m (Ag⁺) + λ^∞_m (Cl^–)

= 61.92 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} + 76.34 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} = 138.26 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

K = 1.826 \times 10^{6} {ohm}^{- 1} {cm}^{2}

Solubility (in mol L^-1) =

\frac{K \times 1000 {cm}^{3} L^{- 1}}{A_{m}^{\infty}}

= \frac{(1.826 \times 10^{- 6} {ohm}^{- 1}) \times (1000 {cm}^{3} L^{- 1})}{(138.26 {ohm}^{- 1} {cm}^{2} {mol}^{- 1})} = 1.32 \times 10^{- 5} mol L^{- 1} .

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