The limiting molar conductivities \(\Lambda ^o\) for NaCl, KBr and KCl are 126, 152 and 150 S cm² mol^–1 respectively. The \(\Lambda ^o\) for NaBr (S cm² mol^–1) is

128

Given, the molar conductivities are

\(\Lambda^o_{NaCl} = 126 Scm^2mol^{-1}\)

\(\Lambda^o_{KBr} = 152 Scm^2mol^{-1}\)

\(\Lambda^o_{KCl} = 150 Scm^2mol^{-1}\)

According to Kohlrausch's law, we can write

 \(\Lambda^o_{NaCl} = \Lambda ^o_{Na^+} + \Lambda ^o_{Cl^-}\) --------(1)

\(\Lambda^o_{KBr} = \Lambda ^o_{K^+} + \Lambda ^o_{Br^-}\) --------(2)

\(\Lambda^o_{KCl} = \Lambda ^o_{K^+} + \Lambda ^o_{Cl^-}\) --------(3)

\(\Lambda^o_{NaBr} = \Lambda ^o_{Na^+} + \Lambda ^o_{Br^-}\) --------(4)

Equation (1) + (2) - (3), we get

\(\Lambda^o_{NaCl} + \Lambda^o_{KBr} − \Lambda^o_{NaBr} = \Lambda ^o_{Na^+} + \Lambda ^o_{Cl^-} + \Lambda ^o_{K^+} + \Lambda ^o_{Br^-} − \Lambda ^o_{K^+} − \Lambda ^o_{Cl^-}\)

⇒ \(126 + 152 − 150 = \Lambda ^o_{Na^+} + \Lambda ^o_{Br^-}\)

⇒\(\Lambda^o_{NaBr} = 128 Scm^2mol^{-1}\)