The resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution of 0.5 M solution of the same electrolyte is 1.4 S m^-1 and resistance of the same solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in Sm^-2 mol^-1 is

5 x 10^-4

5 x 10^-3

5 x 10³

5 x 10²

Solution

5 x 10^-4

For first solution,
k = 1.4 Sm^-1, R =50 Ω , M =0.2
specific conductance
$straight k space equals space 1 over straight R space straight x l over straight A 1.4 space Sm to the power of negative 1 end exponent space equals space 1 over 50 straight x l over straight A l over straight A space equals space 50 space straight x space 1.4 space straight m to the power of negative 1 end exponent straight k space equals 1 over 280 space straight x space 1.4 space straight x space 50 space equals space 1 fourth Now comma space molar space conductivity straight lambda subscript straight m space equals space fraction numerator straight k over denominator 1000 space xm end fraction space equals space fraction numerator 1 divided by 4 over denominator 1000 straight x 0.5 end fraction space equals space 1 over 2000 space equals space 5 space straight x space 10 to the power of negative 4 end exponent space Sm squared mol to the power of negative 1 end exponent$