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  • How to solve this problem- - Redox Reaction and Electrochemistry - JEE Main

Resistance of a conductivity cell all filled with 0.2 M NaCl is 100 \Omega. If the conductance of the same cell when filled with 0.02 M NaCl solution is 200 \Omega, hat is the molar conductivity of 0.02 M NaCl solution if conductivity of 0.2M NaCl is 1.29 Sm-1?

  • Option 1)

    297 S cm2 mol-1

  • Option 2)

    278 S cm2 mol-1

  • Option 3)

    322 S cm2 mol-1

  • Option 4)

    300 S cm2 mol-1

 

Answers (1)

As we have learnt,

 

Formula of Molar Conductivity -

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

-

 

 Cell constant  = 100 x 1.29 = 129 m-1

For 0.02 M solution, conductivity = \frac{129}{200} = 0.645 = 6.45\times 10^{-2}S\:cm^{-1}

Molar conductivity = 6.45\times 10^{-2}S\:cm^{-1} \times \frac{1000}{0.02} cm^3 mol^{-1} =322.5S\:cm^2\:mol^{-1}

 


Option 1)

297 S cm2 mol-1

Option 2)

278 S cm2 mol-1

Option 3)

322 S cm2 mol-1

Option 4)

300 S cm2 mol-1

Posted by

Vakul

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