JEE Advanced levelSingle correct optionDifficulty 3 of 3ec181
A conductivity cell shows a resistance of 500Ω when filled with 0.0100M KCl (κ=1.41×10−3Scm−1). The same cell is then filled with a 0.100M solution of a salt whose molar conductivity at this concentration is 110Scm2mol−1. The resistance now measured is closest to
A64Ω
B0.064Ω
C16Ω
D640Ω
Answer: A
Cell constant G∗=κR=1.41×10−3×500=0.705cm−1. Conductivity of the salt solution: κ=Λmc/1000=110×0.100/1000=1.10×10−2Scm−1. Resistance R=G∗/κ=0.705/0.0110=64.1Ω. Forgetting the factor 1000 gives 0.064Ω.