Resistance of 0.2 M solution of an electrolyte is 50Ω.The specific conductance of the solution is 1.4 Sm−1. The resistance of 0.5 M solution of the same electrolyte is 280Ω. The molar conductivity of 0.5 M solution of the electrolyte in Sm2mol−1 is
In order to solve the problem, calculate the value of cell constant of the first solution and then use this value of cell constant to calculate the value of k of second solution. Afterwards, finally calculate molar conductivity using value of k and m.
For first solution,
k=1.4Sm−1,R=50Ω,M=0.2
Specific conductance (K)=1R×lA
1.4 Sm−1=150×lA⇒lA=50×1.4m−1
For second solution, R=280,lA=50×1.4m−1
K=1280×1.4×50=14
Now, molar conductivity
λm=K1000×m=141000×0.5=12000=5×10−4Sm2mol−1