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  • Solve this problem - Redox Reaction and Electrochemistry - JEE Main

If molar conductance values of Ca2+ and Cl- at \infty dilution are respectively 118.88 \times 10^{-4} m2 mho mol-1 and 77.33\times 10^{-4}m2 mho mol-1  then that of CaCl2 is : 

  • Option 1)

     11.8.88\times 10^{-4}

  • Option 2)

    154.66\times 10^{-4}

  • Option 3)

    273.54\times 10^{-4}

  • Option 4)

    196.22\times 10^{-4}

 

Answers (1)

best_answer

As we have learnt,

 

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_{m\;CaCl_2}^o = \lambda^o_{Ca^{2+}} + 2\lambda^o_{Cl^-} = 118.88\times 10^{-4} + 2(77.33\times 10^{-4}) = 273.54\times 10^{-4} m^2mho\;mol^{-1}

 


Option 1)

 11.8.88\times 10^{-4}

Option 2)

154.66\times 10^{-4}

Option 3)

273.54\times 10^{-4}

Option 4)

196.22\times 10^{-4}

Posted by

satyajeet

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