ElectrochemistryHard

Question

The conductivity of a solution of AgCl at 298 K is found to be 1.382 × 10-6 W-1 cm-1. The ionic conductance of Ag+ and Cl- at infinite dilution are 61.9 W-1 cm2 mol-1 and  76.3 W-1 cm2 mol-1, respecitvley. The solubility of AgCl is

Options

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

Solution

K = 1.382 × 10-6 s cm-1
ΛAgCl = 61.9 + 76.3 = 138.2 =
S = 10-5 M.

Create a free account to view solution

View Solution Free
Topic: Electrochemistry·Practice all Electrochemistry questions

More Electrochemistry Questions