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) $248\ S\ {cm}^{2}\ {mol}^{- 1}$
C) $328\ S\ {cm}^{2}\ {mol}^{- 1}$
D) $348\ S\ {cm}^{2}\ {mol}^{- 1}$
Explanation
(a)
$\left( 126\ {Scm}^{2}\ {mol}^{- 1} \right)\Lambda_{NaCl}^{\infty} = \ \Lambda_{{Na}^{+}}^{\infty} + \ \Lambda_{{Cl}^{-}}^{\infty}\ \ \ \ \$…(i)
$\left( 152\ {Scm}^{2}\ {mol}^{- 1} \right)\Lambda_{KBr}^{\infty} = \ \Lambda_{K^{+}}^{\infty} + \ \Lambda_{{Br}^{-}}^{\infty}\ \ \ \ \ \ \$…(ii)
$\left( 150\ {Scm}^{2}\ {mol}^{- 1} \right)\Lambda_{KCl}^{\infty} = \ \Lambda_{K^{+}}^{\infty} + \ \Lambda_{{Cl}^{-}}^{\infty}\ \ \ \ \ \ \ \$…(iii)
By Eqs. (i) + (ii) – (iii)
$\because\ \Lambda_{NaBr}^{\infty} = \ \Lambda_{{Na}^{+}}^{\infty} + \ \Lambda_{{Br}^{-}}^{\infty}$
$= 126 + 152 - 150$
$= 128\ {Scm}^{2}\ {mol}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.