Question

Limiting molar conductivity of $NaBr$ is

• A. ${\Lambda }_m^{\circ }NaBr={\Lambda }_m^{\circ }NaCl+{\Lambda }_m^{\circ }KBr$
• B. ${\Lambda }_m^{\circ }NaBr={\Lambda }_m^{\circ }NaCl+{\Lambda }_m^{\circ }KBr-{\Lambda }_m^{\circ }KCl$
• C. ${\Lambda }_m^{\circ }NaBr={\Lambda }_m^{\circ }NaOH+{\Lambda }_m^{\circ }NaBr-{\Lambda }_m^{\circ }NaCl$
• D. ${\Lambda }_m^{\circ }NaBr={\Lambda }_m^{\circ }NaCl+{\Lambda }_m^{\circ }NaBr$

Answer 1

Answer: (B)

We can obtain the Limiting molar conductivity of a electrolyte by combination of other electrolytes. All we have to do is to get the correct ionic forms of the required electrolyte. Here for, $NaBr=N{a}^{+}+B{r}^{-}=N{a}^{+}+C{l}^{-}+K^{+}+B{r}^{-}-K^{+}-C{l}^{-}$ Hence option B given this.
${\Lambda }_m^{0}NaBr={\Lambda }_m^{0}NaCl+{\Lambda }_m^{0}KBr-{\Lambda }_m^{0}KCl$