Given that ∧m∞=133.4AgNO3∧m∞=149.9(KCl);∧m∞=144.9Scm2mol−1KNO3  the molar conductivity at infinite dilution for AgCl is:

Λm∞  AgNO3  =  Λm∞  Ag+  +   Λm∞  NO3− = 133.4  sm2  mole−1      ..........(1)Λm∞  KCl  =  Λm∞  K+  +   Λm∞  Cl − = 149.9  sm2  mole−1                           ............(2)Λm∞  KNO3  =  Λm∞  K+  +   Λm∞  NO3− = 144.9  sm2  mole−1              ............(3) Λm∞  AgCl  =  Eq1  + Eq2  −  Eq3= (133.4 + 149.9 – 144.9)Scm2 mole–1        = 138.4 Scm2 mole–1