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

For 0.2 M solution, Specific conductance (κ)=Conductance (1R)×Cell constant (l/a) ⇒κ=150×la=1.3 S m−1⇒la=1.3×50 m−1la=1.3×50×10−2 cm−1For 0.4 M solution, R=260 ΩR=ρla⇒1ρ=κ=1R×laκ=1260×1.3×50×10−2 S cm−1Molar conductivity of the solution is given by Λm=κ×1000M=1260×1.3×50×10−2×10004=6.25 S cm2 mol−1=6.25×10−4 S m2 mol−1