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Electrochemistry question

2023 · Q87

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Chemistry
Electrochemistry
2023 · Q87

Electrochemistry question

2023 · Q87

NEETChemistryElectrochemistryMCQ+4 / −1

The conductivity of centimolar solution of $\mathrm{KCl}$ at $25^{\circ} \mathrm{C}$ is $0.0210 ~\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$ and the resistance of the cell containing the solution at $25^{\circ} \mathrm{C}$ is $60 ~\mathrm{ohm}$ . The value of cell constant is -

A
$3.28 \mathrm{~cm}^{-1}$
B
$1.26 \mathrm{~cm}^{-1}$
C
$3.34 \mathrm{~cm}^{-1}$
D
$1.34 \mathrm{~cm}^{-1}$

View written solutionFree

Correct answer: B

Centimolar solution $$=\frac{1}{100} \mathrm{M}=0.01 ~\mathrm{M}$$

Conductivity $$(k)=0.0210 ~\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$$

Resistance $$(\mathrm{R})=60 ~\mathrm{ohm}$$

$$k=\frac{1}{\mathrm{R}}\left(\frac{\ell}{\mathrm{A}}\right)$$

$$\Rightarrow 0.0210=\frac{1}{60}\left(\frac{\ell}{\mathrm{A}}\right) \Rightarrow \frac{\ell}{\mathrm{A}}=1.26 \mathrm{~cm}^{-1}$$

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