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Q.

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

44.44 Ω

b

33.33 Ω

c

11.11 Ω

d

22.22 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−1

R=1κ(ℓA)=1(0.01 S cm−1)(13cm−1)=33.33 Ω

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