Question

The limiting molar conductivities ${\Lambda }^0$
for $NaCl,KBr\text{and}KCl$ are $126,152\text{and}150Sc{m}^{2}mo{l}^{-1}$ respectively. The ${\Lambda }^0$ for $NaBr$ is:

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

Answer 1

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


For a strong electrolyte $NaBr$, the limiting molar conductivity is the sum of the individual ionic conductivities.

${\Lambda }_{NaBr}^0={\lambda }_{N{a}^{+}}^0+{\lambda }_{B{r}^{-}}^0$

${\Lambda }_{NaBr}^0={\Lambda }_{NaCl}^0+{\Lambda }_{KBr}^0-{\Lambda }_{KCl}^0....eqn\left(1\right)$

${\Lambda }_{NaBr}^0={\lambda }_{N{a}^{+}}^0+{\lambda }_{C{l}^{-}}^0+{\lambda }_{K^{+}}^0+{\lambda }_{B{r}^{-}}^0-{\lambda }_{N{a}^{+}}^0-{\lambda }_{B{r}^{-}}^0$

${\Lambda }_{NaBr}^0={\lambda }_{N{a}^{+}}^0+{\lambda }_{B{r}^{-}}^0$

Hence,
Substituting the values in equation (1), we get-

${\Lambda }_{NaBr}^0={\Lambda }_{NaC1}^0+{\Lambda }_{KBr}^0-{\Lambda }_{KC1}^0$

${\Lambda }_{NaBr}^0=126+152-150$
${\Lambda }_{NaBr}^0=128Sc{m}^{2}mo{l}^{-1}$

Hence, option A is correct.