The molar conductivity of 0.05 M of solution of an electrolyte is 200 Ω−1 cm2 mol−1. The resistance offered by a conductivity cell with cell constant (1/3) cm−1 would be about
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a
11.11 Ω
b
22.22 Ω
c
33.33 Ω
d
44.44 Ω
answer is C.
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Detailed Solution
κ=Λc=(200 S cm2 mol−1)(0.05×10−3 mol cm−1)=0.01 S cm−1R=1κ(ℓA)=1(0.01 S cm−1)(13cm−1)=33.33 Ω