When a certain cell was filled with 0.02N KCl solution, the resistance was 350ohm when the same cell was filled with 0.01N eq solution of sodium acetate the resistance was 1158ohm. If the specific conducance of 0.02N KCl solution is 27.7*10^-4ohm^-1cm^-1. calculate equivalent conductance of sodium acetate solution in ohm^-1cm^-1gequiv^-1?
Solution) GIVEN concentration of KCl = 0.02
resistance =350ohm
sodium acetate the resistance = 1158ohms
equivalent conductance =27.7*10^-4ohm-1cm-1.
k=ρl/a
k/ρ=(27.7*10*10^-4)/350
k of sodium acetate= l/a*1/R
equvalent conductance =(k*1000)/.02