[Four-digit Integer] The specific conductivity of a saturated solution of AgCl is 2.80 × 10^-4 mho m — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Step 1: Calculate molar conductance at infinite dilution.} \[ \lambda^\circ_m(\text{AgCl}) = \lambda^\circ(\text{Ag}^+) + \lambda^\circ(\text{Cl}^-) = 6.19 \times 10^{-3} + 7.81 \times 10^{-3} = 1.40 \times 10^{-2}\ \text{mho m}^2\text{mol}^{-1} \] \textbf{Step 2: Calculate molar solubility.} \[ C = \frac{\kappa}{\lambda^\circ_m} = \frac{2.80 \times 10^{-4}}{1.40 \times 10^{-2}} = 2.0 \times 10^{-2}\ \text{mol m}^{-3} = 2.0 \times 10^{-5}\ \text{mol L}^{-1} \] \textbf{Step 3: Convert to g L⁻¹.} $M(\text{AgCl}) = 108 + 35.5 = 143.5\ \text{g mol}^{-1}$ \[ \text{Solubility} = 2.0 \times 10^{-5} \times 143.5 = 2.87 \times 10^{-3}\ \text{g L}^{-1} \] \textbf{Step 4: Express in order of 10⁻⁵ g L⁻¹.} \[ 2.87 \times 10^{-3}\ \text{g L}^{-1} = 287 \times 10^{-5}\ \text{g L}^{-1} \] \[ \boxed{0287} \]