-->

Electrochemistry

Question
CBSEENCH12010870

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω. 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 Ω.The molar conductivity of 0.02 M solution of the electrolyte will be

  • 124 × 10–4 Sm2 mol–1

  • 1240 × 10–4 Sm2 mol–1

  • 1.24 × 10–4 Sm2 mol–1

  • 12.4 × 10–4 Sm2 mol–1

Solution

D.

12.4 × 10–4 Sm2 mol–1

straight R space equals space 100 space straight capital omega
straight K space equals space 1 over straight R open parentheses calligraphic l over straight a close parentheses
1.29 space equals space 1 over 100 open parentheses calligraphic l over straight a close parentheses
open parentheses calligraphic l over straight a close parentheses space equals space 129 space straight m to the power of negative 1 end exponent
space straight R space equals space 520 space space straight capital omega comma space straight C space equals space 0.2 space straight M
straight K space equals space 1 over straight R open parentheses calligraphic l over straight a close parentheses space equals space 1 over 520 space left parenthesis 129 right parenthesis space straight capital omega to the power of negative 1 end exponent straight m to the power of negative 1 end exponent
straight mu space equals space straight K space straight x space straight V subscript in space cm cubed
space equals space 1 over 520 space straight x space 129 space straight x space fraction numerator 1000 over denominator 0.2 end fraction space straight x space 10 to the power of negative 6 end exponent straight m cubed
space equals space 129 over 520 space straight x space fraction numerator 1000 over denominator 0.2 end fraction space straight x space 10 to the power of negative 6 end exponent
space equals space 1.24 space straight x space 10 to the power of negative 3 end exponent
equals 12.4 space straight x space 10 to the power of negative 4 end exponent