Question

The limiting molar conductivities ${\wedge }^0$ for $NaCl,KBr$ and $KCl$ are $126,152$ and $150$ $S$ $c{m}^{2}mo{l}^{-1}$ respectively. The ${\wedge }^0$ 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

Answer: (A)

$128Sc{m}^{2}mo{l}^{-1}$

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

${\Lambda }_{NaBr}^0={\lambda }_{Na+}^0{+\lambda }_{Br-}^0$

Hence, ${\Lambda }_{NaBr}^0={\Lambda }_{NaCl}^0+{\Lambda }_{KBr}^0-{\Lambda }_{KCl}^0$
Substituting values, we get-

${\Lambda }_{NaBr}^0={\Lambda }_{NaCl}^0+{\Lambda }_{KBr}^0-{\Lambda }_{KCl}^0=(126+152-150)S{cm}^2{mol}^{-1}=128S{cm}^2{mol}^{-1}.$

Hence, option A is correct.