The limiting molar conductivities \Lambda ^{\circ}for NaCl,KBr \:\; and \; \: KCl\, are\, 126,152\, and\, 150\, S\, cm^{2}\, mol^{-1}\, respectively.

\, The\, \Lambda ^{\circ}\, for\, NaBr\, is

  • Option 1)

    128\; S\; cm^{2}\; mol^{-1}

  • Option 2)

    176\; S\; cm^{2}\; mol^{-1}

  • Option 3)

    278\; S\; cm^{2}\; mol^{-1}

  • Option 4)

    302\; S\; cm^{2}\; mol^{-1}

 

Answers (1)

As we learnt in

Formula for limiting molar conductivity for electrolyte -

\Lambda_{m}^{0}= \lambda _{+}^{0}.\nu _{+}\:+\:\nu _{-}.\lambda_{-}^{0}

\lambda _{+}^{0}\:and\:\lambda_{-}^{0}

are limiting molar conductivities of cation and anion respectively.

- wherein

if an electrolyte on dissociation gives \nu_{+} \:cation\:and\: \nu_{-} anions.

 

 

\Lambda ^{\circ}_{NaCl}=\Lambda ^{\circ}_{Na^{+}}+\Lambda ^{\circ}_{Cl^{-}}\; \; \; .........(i)

\Lambda ^{\circ}_{KBr}=\Lambda ^{\circ}_{K^{+}}+\Lambda ^{\circ}_{Br^{-}}\; \; \; .........(ii)

\Lambda ^{\circ}_{KCl}=\Lambda ^{\circ}_{K^{+}}+\Lambda ^{\circ}_{Cl^{-}}\; \; \; .........(iii)

Equation (i) + (ii) – (iii)

\Lambda ^{\circ}_{NaBr}=\Lambda ^{\circ}_{Na^{+}}+\Lambda ^{\circ}_{Br^{-}}\; \; \;

=126+152-150=128\, S\, cm^{2}\, mol^{-1}


Option 1)

128\; S\; cm^{2}\; mol^{-1}

This option is correct.

Option 2)

176\; S\; cm^{2}\; mol^{-1}

This option is incorrect.

Option 3)

278\; S\; cm^{2}\; mol^{-1}

This option is incorrect.

Option 4)

302\; S\; cm^{2}\; mol^{-1}

This option is incorrect.

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