Question

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

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

Answer 1

Answer: (A)

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