Question

Limiting molar conductivity of $NaBr$ is

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

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, $NaBr = Na ^{+}+ Br ^{-}= Na ^{+}+ Cl ^{-}+ K ^{+}+ Br ^{-}- K ^{+}- Cl ^{-}$ Hence option B given this.
$\Lambda_{ m }^{0} NaBr =\Lambda_{ m }^{0} NaCl +\Lambda_{ m }^{0} KBr -\Lambda_{ m }^{0} KCl$