The conductivity of a solution of AgCl at 298 K is found to be 1.382 × 10–6 Ω–1 cm–1. The ionic conductance of Ag+ and Cl– at infinite dilution are 61.9 Ω–1 cm2 mol–1 and 76.3 Ω–1 cm2 mol–1, respectively. The solubility of AgCl is (A) 1.4 × 10–5 mol L–1 (b) 1 × 10–2 mol L–1 (c) 1 × 10–5 mol L–1 (d) 1.9 × 10–5 mol L–1 Share with your friends Share 17 Vartika Jain answered this Dear Student, The molar conductivity at infinite dilution, λ0 is the sum of those of ions. λ0 = λ0(Ag+)+λ0(Cl-) =61.9 +76.3 =138.2 Ω-1cm2mol-1and, λ0=κc138.2 Ω-1cm2mol-1 =1.382×10-6 Ω-1cm-1c or, c= 1.382×10-6 Ω-1cm-1138.2 Ω-1cm2mol-1 =1×10-8 molcm-3Now, 1 molcm-3= 103 molL-1Therefore, c=1×10-8 molcm-3=1×10-5 molL-1Hence, correct answer is (C) 13 View Full Answer