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Chemistry Question on Electrochemistry

Limiting molar conductivity of NaBrNaBr is

A

Λm∘NaBr=Λm∘NaCl+Λm∘KBr\Lambda^{\circ}_{m}NaBr=\Lambda^{\circ}_{m}NaCl +\Lambda^{\circ}_{m}KBr

B

Λm∘NaBr=Λm∘NaCl+Λm∘KBr−Λm∘KCl\Lambda^{\circ}_{m}NaBr=\Lambda^{\circ}_{m}NaCl +\Lambda^{\circ}_{m}KBr-\Lambda^{\circ}_{m}KCl

C

Λm∘NaBr=Λm∘NaOH+Λm∘NaBr−Λm∘NaCl\Lambda^{\circ}_{m}NaBr=\Lambda^{\circ}_{m}NaOH +\Lambda^{\circ}_{m}NaBr-\Lambda^{\circ}_{m}NaCl

D

Λm∘NaBr=Λm∘NaCl+Λm∘NaBr\Lambda^{\circ}_{m}NaBr=\Lambda^{\circ}_{m}NaCl +\Lambda^{\circ}_{m}NaBr

Answer

Λm∘NaBr=Λm∘NaCl+Λm∘KBr−Λm∘KCl\Lambda^{\circ}_{m}NaBr=\Lambda^{\circ}_{m}NaCl +\Lambda^{\circ}_{m}KBr-\Lambda^{\circ}_{m}KCl

Explanation

Solution

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−NaBr = Na ^{+}+ Br ^{-}= Na ^{+}+ Cl ^{-}+ K ^{+}+ Br ^{-}- K ^{+}- Cl ^{-} Hence option B given this. Λm0NaBr=Λm0NaCl+Λm0KBr−Λm0KCl\Lambda_{ m }^{0} NaBr =\Lambda_{ m }^{0} NaCl +\Lambda_{ m }^{0} KBr -\Lambda_{ m }^{0} KCl