Resistance of 0.2 M solution of an electrolyte is 50Ω. The specific conductance of the solution is 1.3 Sm-1.If resistance of the 0.4 M solution of the same electrolyte is 260Ω,its molar conductivity is

Specific conductance = conductance x cell constant              1.3Sm−1=150S× cell constant ∴  Cell constant =1.3×50m−1=65 m−1=(65/100)cm−1 Molar conductivity =1000× conductance × cell constant  molarity                                  =10000.4×1260×65100=6.25Scm2mol−1=6.25×10−4Sm2mol−1