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) 10 Ω-1 cm 2 g equiv -1 (B) 20 Ω-1 cm 2 g equiv -1 (C) 300 Ω-1 cm 2 g equiv -1 (D) 100 Ω-1 cm 2 g equiv -1

Tardigrade Admin · Accepted Answer

Given, (l/ a )=0.5 cm -1 R =50 Ω N =1.0 Specific conductance ( k )=(1/ρ)=(l/ R ⋅ a )=(0.5/50) Lambda= k × (1000/ N )=(0.5/50) × (1000/0.1) =10 Ω-1 cm 2 g 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

$10 Ω^{- 1} c m^{2} g$ equiv $^{- 1}$

$20 Ω^{- 1} c m^{2} g$ equiv $^{- 1}$

$300 Ω^{- 1} c m^{2} g$ equiv $^{- 1}$

$100 Ω^{- 1} c m^{2} g$ equiv $^{- 1}$

Given,
$\frac{l}{a} = 0.5 c m^{- 1}$
$R = 50 Ω$
$N = 1.0$
Specific conductance
$(k) = \frac{1}{ρ} = \frac{l}{R \cdot a} = \frac{0.5}{50}$
$Λ = k \times \frac{1000}{N} = \frac{0.5}{50} \times \frac{1000}{0.1}$
$= 10 Ω^{- 1} c m^{2} g e q^{- 1}$

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

10Ω−1cm2g equiv −1

20Ω−1cm2g equiv −1

300Ω−1cm2g equiv −1

100Ω−1cm2g equiv −1