Question
Chemistry Question on The solid state
The conductivity of 0.02 N solution of a cell of KCl at 25∘C is 2.765×10−3S cm−1. If the resistance of a cell containing this solution is 4×103 ohm, what will be the cell constant?
Answer
Explanation:
Given:Normality concentration of KCl solution =0.02NConductivity of KCl solution K=2.765×10−3Scm−1Temperature, T=25∘CResistance of the cell, R=4000 OhmWe have to find the value of the cell constant.Conductivity, k is given as:κ=1R[1A]where, K¯= ConductivityR= ResistanceI= DistanceA= Cross section Area1A= Cell constantThus, from equation (i) 1A=2.765×10−3×4×103=2.765×10−3Scm−1×4×103S−1=2.765×4 cm−1=11.06 cm−1Therefore cell constant is 11.06 cm−1.Hence, the correct answer is 11.06 cm−1.