Question

Limiting molar conductivity of NaBr is:

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

Answer 1

The correct option is B ${\Lambda }_m^{o}NaBr={\Lambda }_m^{o}NaCl+{\Lambda }_m^{o}KBr-{\Lambda }_m^{o}KCl$

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.

${\Lambda }_m^{0}NaBr={\Lambda }_m^{0}NaCl+{\Lambda }_m^{0}KBr-{\Lambda }_m^{0}KCl$