Verified Solution • 2025–26 Session moderate

QJEEM06Chde01: JEE Main 2006 Chemistry - Class 12 JEEM06Chde01 Solution: Resistance of a conductivity cell filled with a solution of an…

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω\Omega . The conductivity of this solution is 1.29 S m–1. Resistance of the same cell when filled with 0.2 M of the same solution is 520 Ω\Omega , The molar conductivity of 0.02 M solution of the electrolyte will be

Multiple Choice Options

  • A 124 ×\times 10–4 S m2 mol–1
  • B 1240 ×\times 10–4 S m2 mol–1
  • C 1.24 ×\times 10–4 S m2 mol–1
  • D 12.4 ×\times 10–4 S m2 mol–1

Step-by-Step Verified Solution

R=100Ω,κ=1R(la),laR = 100\Omega ,\kappa = {1 \over R}\left( {{l \over a}} \right),{l \over a} (cell constant)

                 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,                  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =1.29×100m−1= 1.29 \times 100{m^{ - 1}}

Given, R=520Ω,C=0.2M, μR = 520\Omega ,C = 0.2M,\,\mu (molar conductivity) == ?

μ=κ×V        \mu = \kappa \times V\,\,\,\,\,\,\,\, ( κ\kappa can be calculated as κ=1R(1a)\kappa = {1 \over R}\left( {{1 \over a}} \right)

now cell constant is known.

Hence, μ=1520×129×10000.2×10−6m3\mu = {1 \over {520}} \times 129 \times {{1000} \over {0.2}} \times {10^{ - 6}}{m^3}

=12.4×10−4 Sm2mol−1= 12.4 \times {10^{ - 4}}\,S{m^2}mo{l^{ - 1}}

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