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Electrochemistry

Question
CBSEENCH12010127
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The resistance of a conductivity cell filled with 0.1 mol L-1 KCl solution is 100  . If the resistance of the same cell when filled with 0.02 mol L-1 KCl solution is 520 straight capital omega  , calculate the conductivity and molar conductivity of 0.02 mol L-1  KCl solution. The conductivity of 0.1 mol L-1  KCl solution is 1.29x 10-2 straight capital omega -1cm-1

Solution

Given that:

Concentration of the KCl solution = 0.1 mol L-1

Resistance of cell filled with 0.1 mol L-1 KCl solution = 100 ohm

Cell constant = G* = conductivity x resistance

1.29x10-2 ohm-1 cm-1 x 100 ohm = 1.29 cm-1 = 129 m-1

Cell constant for a particular conductivity cell is a constant.

Conductivity of 0.02 mol L-1  KCl solution =  =0.248 Sm-1

Concentration = 0.02 mol-1 


                     = 1000x 0.02 mol m-3 = 20 mol m-3


Now,

Molar conductivity = 
straight lambda subscript straight m space equals straight k over straight c equals fraction numerator 248 space straight x 10 to the power of negative 3 end exponent space Sm to the power of negative 1 end exponent over denominator 20 space mol space straight m to the power of negative straight s end exponent end fraction space equals 124 space straight x space 10 to the power of negative 4 end exponent space straight S space straight m squared space mol to the power of negative 1 end exponent Therefore comma space the space molar space conductivity space of space 0.02 space mol to the power of negative 1 end exponent space KCl space solution space is space 124 space straight x space 10 to the power of negative 4 end exponent space straight S space straight m squared space mol to the power of negative 1 end exponent


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