The limiting molar conductivities Λ0
for NaCl, KBr and KCl are 126, 152 and 150 Scm2mol−1 respectively. The Λ0 for NaBr is:
Λ0NaBr=λ0Na++λ0Br−
Λ0NaBr=Λ0NaCl+Λ0KBr−Λ0KCl....eqn(1)
Λ0NaBr=λ0Na++λ0Cl−+λ0K++λ0Br−−λ0Na+−λ0Br−
Λ0NaBr=λ0Na++λ0Br−
Hence,
Substituting the values in equation (1), we get-
Λ0NaBr=Λ0NaC1+Λ0KBr−Λ0KC1
Λ0NaBr=126+152−150
Λ0NaBr=128 Scm2mol−1
Hence, option A is correct.