SATHEE: Redox Reactions And Electrochemistry Ques 328

Redox Reactions And Electrochemistry Ques 328

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

Options:

A) $100, o h m^{- 1} c m^{2} g, E q^{- 1}$

B) $20, o h m^{- 1} c m^{2} g, E q^{- 1}$

C) $300, o h m^{- 1} c m^{2} g, E q^{- 1}$

D) $10, o h m^{- 1} c m^{2} g, E q^{- 1}$

Show Answer

Answer:

Correct Answer: D

Solution:

[d] $l / a = 0.5, c m^{- 1}, R = 50, Ω$ $ρ = \frac{R a}{l} = \frac{50}{0.5} = 100$

$A = k \times \frac{1000}{N} = \frac{1}{ρ} \times \frac{1000}{N} = \frac{1}{100} \times \frac{1000}{1}$

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

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Redox Reactions And Electrochemistry Ques 328

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

Options:

Answer:

Solution:

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