Question

The limiting molar conductivities $\Lambda^{\circ}$ for NaCl, KBr and KCl are 126, 152 and 150 S $cm^2\, mol^{-1}$ respectively. The $\Lambda^{\circ}$ for NaBr is

• A. $128\, S \,cm^2\, mol^{-1}$
• B. $302\, S \,cm^2\, mol^{-1}$
• C. $278\, S \,cm^2\, mol^{-1}$
• D. $176\, S \,cm^2\, mol^{-1}$

Answer 1

Answer: (A)

$\Lambda^{\circ}_{NaCl} = \lambda^{\circ}_{Na} +\lambda^{\circ}_{Cl} = 126\quad\dots \left(1\right)$
$Λ^{\circ}_{KBr} = λ^{\circ}_{K^{+}} + λ^{\circ}_{Br^{-}} = 152\quad\quad\dots \left(2\right)$
$Λ^{\circ}_{KCl} = λ^{\circ}_{K^{+}} + λ^{\circ}_{Br^{-}} = 150\quad\quad\dots \left(3\right)$
$\Lambda^{\circ}_{NaBr} = λ^{\circ}_{Na} + λ^{\circ}_{Br^{-}}$
$\Lambda ^{\circ }_{NaBr} = 126 + 152 −150 = 128$