The conductivity of a saturated solution of AgCl at 298 K was found to be 3.40 × 10^–6 Ω^–1 cm^–1; the conductivity of water used to make up the solution was 1.60 × 10^–6 Ω^–1 cm^–1. Determine the sol

Question

The conductivity of a saturated solution of AgCl at 298 K was found to be 3.40 × 10^–6 Ω^–1 cm^–1; the conductivity of water used to make up the solution was 1.60 × 10^–6 Ω^–1 cm^–1. Determine the solubility of AgCl in water in mole per litre at 298 K. The equivalent conductivity of AgCl at infinite dilution is 150.0 Ω^–1 cm^2 eq^–1.

Prashant Kotian · Accepted Answer

Correct answer: B. Specific conductance due to dissolved AgCl (κ) = κ_solution - κ_water = 3.40 × 10^–6 - 1.60 × 10^–6 = 1.80 × 10^–6 Ω^–1 cm^–1. Solubility S = (κ × 1000) / Λ_m. Since the valency is 1, equivalent conductivity equals molar conductivity (150.0). S = (1.80 × 10^–6 × 1000) / 150.0 = 1.80 × 10^–3 / 150.0 = 1.2 × 10^–5 M. Therefore, correct answer is B.

💡 Solution & Explanation

Practice 22,000+ questions like this