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

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) 

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