Given l / a=0.5 cm -1, R=50 ohm, N=1.0. The equivalent conductance of the electrolytic cell is

Question

Given l / a=0.5 cm -1, R=50 ohm, N=1.0. The equivalent conductance of the electrolytic cell is (A) 100 ohm -1 cm -2 gm eq -1 (B) 20 ohm -1 cm 2 gm eq -1 (C) 300 ohm -1 cm 2 gm eq -1 (D) 10 ohm -1 cm 2 gm eq -1

Tardigrade Admin · Accepted Answer

l l a=0.5 cm -1, R=50 ohm p=(R a/l)=(50/0.5)=100 Lambda= kappa × (1000/N)=(1/p) × (1000/N)=(1/100) × (1000/1) =10 ohm -1 cm 2 gm eq -1

Q. Given $l / a = 0.5 c m^{- 1}, R = 50 o hm, N = 1.0.$ The equivalent conductance of the electrolytic cell is

$100 o h m^{- 1} c m^{- 2} g m e q^{- 1}$

$20 o h m^{- 1} c m^{2} g m e q^{- 1}$

$300 o h m^{- 1} c m^{2} g m e q^{- 1}$

$10 o h m^{- 1} c m^{2} g m e q^{- 1}$

$ll a = 0.5 c m^{- 1}, R = 50 o hm$

$p = \frac{R a}{l} = \frac{50}{0.5} = 100$

$Λ = κ \times \frac{1000}{N} = \frac{1}{p} \times \frac{1000}{N} = \frac{1}{100} \times \frac{1000}{1}$

$= 10 o h m^{- 1} c m^{2} g m e q^{- 1}$

Q. Given l/a=0.5cm−1,R=50ohm,N=1.0. The equivalent conductance of the electrolytic cell is

100ohm−1cm−2gmeq−1

20ohm−1cm2gmeq−1

300ohm−1cm2gmeq−1

10ohm−1cm2gmeq−1