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Standard XII

Chemistry

Faraday's Second Law

A KCl solutio...

Question

A KCl solution of conductivity $0.14 S m^{- 1}$ shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops 1.03 Ω. The conductivity of the HCl solution is . x $10^{- 2} S m^{- 1}$ . (Round off to the Nearest Integer).

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Solution

For KCl solution,

$R = (\frac{1}{K}) (\frac{l}{A}) \Rightarrow \frac{l}{A} = R \times K = 4.19 \times 0.14$

$\frac{l}{A} = (4.19 \times 0.14) m^{- 1} = 0.58 m^{- 1}$

$\frac{l}{A} is constant$

For $H C l$ solution,

$R = (\frac{1}{K}) (\frac{l}{A})$

$K = (\frac{l / A}{R}) = \frac{0.58}{1.03} = 0.56 = 56 \times 10^{- 2} S m^{- 1}$
The correct answer is 56

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A KCl solution of conductivity 0.14Sm−1 shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops 1.03 Ω. The conductivity of the HCl solution is . x 10−2Sm−1. (Round off to the Nearest Integer).

For KCl solution, R=(1K)(lA)⇒lA=R×K=4.19×0.14 ​ lA=(4.19×0.14) m−1=0.58 m−1 lA is constant For HCl solution, R=(1K)(lA) K=(l/AR)=0.581.03=0.56=56×10−2Sm−1 The correct answer is 56

A KCl solution of conductivity $0.14 S m^{- 1}$ shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops 1.03 Ω. The conductivity of the HCl solution is . x $10^{- 2} S m^{- 1}$ . (Round off to the Nearest Integer).