Volume (in L) of water in the pool is — Electrochemistry Chemistry Question

💡 Solution & Explanation

Total conductivity with NaCl = G* / R = 0.4 / 8000 = 5 × 10^-5 Ω^-1 cm^-1. Conductivity contributed by NaCl (κ_NaCl) = κ_total - κ_water = (5 × 10^-5) - (4 × 10^-5) = 1 × 10^-5 Ω^-1 cm^-1. Concentration of NaCl in Molarity (C) = (κ_NaCl × 1000) / Λ_NaCl = (1 × 10^-5 × 1000) / 125 = 8 × 10^-5 mol/L. Moles of NaCl added = 585 g / 58.5 g/mol = 10 moles. Volume of pool = Moles / Concentration = 10 / (8 × 10^-5) = 125,000 L. Therefore, correct answer is A.