Question

The conductivity of a solution of AgCl at 298 K is found to be $\text{1.382 × 1}\text{0}^{\text{-6}}\ \Omega^{\text{-1}}\text{ c}\text{m}^{\text{-1}}$. The ionic conductance of Ag+ and Cl– at infinite dilution are $\text{61.9}\Omega\ ^{\text{-1}}\text{ c}\text{m}^{2}\text{ mo}\text{l}^{\text{-1}}$ and $\text{76.3 }\Omega\ ^{\text{-1}}\text{ c}\text{m}^{2}\text{ mo}\text{l}^{\text{-1}}\text{ ,}$ respectively. The solubility of AgCl is

• A. 1.4 × 10–5mol L–1
• B. 1 × 10–2mol L–1
• C. 1 × 10–5mol L–1
• D. 1.9 × 10–5mol L–1

Answer 1

Answer: (C)

1 × 10–5mol L–1

Answer 2

K = 1.382 × 10–6 s cm–1

LAgCl = 61.9 + 76.3 = 138.2 = $\frac { 1000 \times 1.382 \times 10 ^ { - 6 } } { S }$

S = 10–5 M.