First slide
Kohlraush law
Question

The limiting molar conductivities λo for NaCl, KBr an KCl are 126, 152 and 150 S. cm2 mol–1  respectively. Thenλo for NaBr is

Easy
Solution

\lambda _{NaBr}^o = \lambda _{NaCl}^o + \lambda _{KBr}^o - \lambda _{KCl}^o = 126 + 152 - 150 = 128Sc{m^2}mo{l^{ - 1}}

 

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