Question

The limiting molar conductivities of $NaCl$, $KBr$ and $KCl$ are 126, 152 and 150 $Sc{m}^{2}mo{l}^{-1}$ respectively. The limiting molar conductivity for $NaBr$ is:

• A. 302 $Sc{m}^{2}mo{l}^{-1}$
• B. 176 $Sc{m}^{2}mo{l}^{-1}$
• C. 278 $Sc{m}^{2}mo{l}^{-1}$
• D. 128 $Sc{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is D 128 $Sc{m}^{2}mo{l}^{-1}$

The limiting molar conductivities of NaCl, KBr and KCl are 126, 152 and 150 $Sc{m}^{2}mo{l}^{-1}$ respectively.

The limiting molar conductivity ${\Lambda }^o$ for NaBr is calculated by the following expression.

${\lambda }_{NaBr}^{\infty }={\lambda }_{NaCl}^{\infty }+{\lambda }_{KBr}^{\infty }-{\lambda }_{KCl}^{\infty }$

${\lambda }_{NaBr}^{\infty }=126+152-150=128Sc{m}^{2}mo{l}^{-1}$