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 cm2 g Eq-1 (B) 20 ohm-1 cm2 g Eq-1 (C) 300 ohm-1 cm2 g Eq-1 (D) 10 ohm-1 cm2 g Eq-1

Tardigrade Admin · Accepted Answer

l/a =0.5 cm-1, R=50Ω ρ =(Ra/l) =(50/0.5)=100 Lambda =k × (1000/N) =(1/ρ) × (1000/N) =(1/100) × (1000/1) =10 ohm-1 Cm2 g Eq-1

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

$100$ ohm $^{- 1} c m^{2} g E q^{- 1}$

$20$ ohm $^{- 1} c m^{2} g E q^{- 1}$

$300$ ohm $^{- 1} c m^{2} g E q^{- 1}$

$10$ ohm $^{- 1} c m^{2} g E q^{- 1}$

$l / a = 0.5 c m^{- 1}, R = 50Ω$
$ρ = \frac{R a}{l} = \frac{50}{0.5} = 100$
$Λ = k \times \frac{1000}{N} = \frac{1}{ρ} \times \frac{1000}{N} = \frac{1}{100} \times \frac{1000}{1}$
$= 10 o h m^{- 1} C m^{2} g E q^{- 1}$

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

100 ohm−1cm2gEq−1

20 ohm−1cm2gEq−1

300 ohm−1cm2gEq−1

10 ohm−1cm2gEq−1