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
Answer & explanation
Correct answer: option 4
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
\(\Lambda^o_{NaCl} + \Lambda^o_{KBr} − \Lambda^o_{NaBr} = \Lambda ^o_{Na^+} + \Lambda ^o_{Cl^-} + \Lambda ^o_{K^+} + \Lambda ^o_{Br^-} − \Lambda ^o_{K^+} − \Lambda ^o_{Cl^-}\)
⇒ \(126 + 152 − 150 = \Lambda ^o_{Na^+} + \Lambda ^o_{Br^-}\)
⇒\(\Lambda^o_{NaBr} = 128 Scm^2mol^{-1}\)