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Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 \Omega The conductivity of this solution is 1.29 Sm-1. Resistance of the same cell when filled with 0.2 M of the same solution is 520 \Omega. The molar conductivity of 0.02 M solution of the electrolyte will be :

  • Option 1)

    124\times 10^{-4} S m^{2} mol^{-1}

  • Option 2)

    1240\times 10^{-4} S m^{2} mol^{-1}

  • Option 3)

    1.24\times 10^{-4} S m^{2} mol^{-1}

  • Option 4)

    12.4\times 10^{-4} S m^{2} mol^{-1}

 

Answers (1)

best_answer

As we learnt in

Molar Conductivity -

\Lambda m = \frac{K}{C}

- wherein

K is expressed in Sm^{-1} and concentration in mol m^{-3} .

Unit of \Lambda m = Sm^{-2} mol^{-1}

 

 and

Formula of Molar Conductivity -

 \Lambda m (S\:cm^{2}mol^{-1})=\frac{\kappa (S\:cm^{-1})}{1000Lm^{-3}\times molarity(mol\:L^{-1})}

-

 

  

\kappa =\frac{1}{R}\left ( \frac{l}{a} \right )\; i.e.,\: 1.29=\frac{1}{100}\left ( \frac{l}{a} \right )

l/a=129\, m^{-1}

R=520\, \Omega \, for\, 0.2M,\, C=0.02\, M

\lambda _{m}=\kappa \times \frac{1000}{molarity}=\frac{1\times 129}{520}\times \frac{1000}{0.02}\times 10^{-6}\, m^{3}

=124\times 10^{-4}\, S\, m^{2}\, mol^{-1}


Option 1)

124\times 10^{-4} S m^{2} mol^{-1}

This option is correct.

Option 2)

1240\times 10^{-4} S m^{2} mol^{-1}

This option is incorrect.

Option 3)

1.24\times 10^{-4} S m^{2} mol^{-1}

This option is incorrect.

Option 4)

12.4\times 10^{-4} S m^{2} mol^{-1}

This option is incorrect.

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