The limiting molar conductivity for KCl, NaBr and NaCl are 100, 120, 130 S cm-2mol-1 respectively. Calculate the limiting molar conductivity for KBr?
90 S cm-2mol-1
We know from Kohlrausch law that for
E0(KCl) = λ0Cl- + λ0K+ = 100 scm-2mol-1
E0(NaBr) = λ0Br- + λ0Na+ = 120 scm-2mol-1
E0(NaCl) = λ0Na+ + λ0Cl- = 130 scm-2mol-1
Subtracting ( iii) from ( i) and add (ii) to it
100 S cm-2mol-1 - 130 S cm-2mol-1 + 120 S cm-2mol-1 = 90 S cm-2mol-1