Q. 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) 176 S ${cm}^{2}{mol}^{- 1}$
C) 278 S ${cm}^{2}{mol}^{- 1}$
D) 302 S ${cm}^{2}{mol}^{- 1}$
Explanation
(a)
By Kohlrausch’s law
$\land_{NaBr}^{{^\circ}} = \ \land_{NaCl}^{{^\circ}}\ + \ \land_{KBr}^{{^\circ}} - \ \land_{KCl}^{{^\circ}}$
= 126 + 152 – 150
$= \ 128\ S\ {cm}^{2}{mol}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.