The specific conductivity of a saturated solution of AgCl is at 25°C. If and , the solubility of sil — Electrochemistry Chemistry Question
Question
The specific conductivity of a saturated solution of AgCl is $2.80 \times 10^{-4} \text{ mho m}^{-1}$ at 25°C. If $\lambda^\circ_{Ag^+} = 6.19 \times 10^{-3} \text{ mho m}^2 \text{ mol}^{-1}$ and $\lambda^\circ_{Cl^-} = 7.81 \text{ mho m}^2 \text{ mol}^{-1}$, the solubility of silver chloride (in order of $10^{-5} \text{ g l}^{-1}$) at 25°C, is
💡 Solution & Explanation
Note the typo in the given unit of $\lambda^\circ_{Cl^-}$; it should match $10^{-3}$ order. Assuming $\lambda^\circ_{Cl^-} = 7.81 \times 10^{-3} \text{ mho m}^2 \text{ mol}^{-1}$, total $\Lambda^\circ_m = (6.19 + 7.81) \times 10^{-3} = 14.0 \times 10^{-3} \text{ S m}^2 \text{ mol}^{-1}$. Concentration $C (\text{mol/m}^3) = \kappa / \Lambda^\circ_m = (2.80 \times 10^{-4}) / (14.0 \times 10^{-3}) = 0.02 \text{ mol/m}^3$. In Molarity, $C = 0.02 / 1000 = 2 \times 10^{-5}$ mol/L. Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol. Solubility = $2 \times 10^{-5} \times 143.5 = 2.87 \times 10^{-3}$ g/L = $287 \times 10^{-5}$ g/L. The required integer in order of $10^{-5}$ is 287. Therefore, correct answer is 0287.