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  • The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 . If the conductivity of 0.0

The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 \Omega. If the conductivity of 0.01M KCl solution at 298 K is \mathrm{0.152 \times 10^{-3} \; S \;cm^{-1}}, then the cell constant of the conductivity cell is  ______ \mathrm{\times 10^{-3} \;cm^{-1}}

Option: 1

266


Option: 2

----------


Option: 3

------------


Option: 4

-------------


Answers (1)

best_answer

We know the formula

\mathrm{ Conductivity(K) = Conductance(G)\: \times \: cell \: constant(G^*)}

\mathrm{K=G \times\left(\frac{\ell}{A}\right)}

\mathrm{\text { Resistance } R=\frac{1}{G}}

Therefore, we can write

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

\mathrm {0.152 \times 10^{-3}=\frac{1}{1750} \times\left(\frac{\ell}{A}\right)}

\mathrm{\left(\frac{\ell}{A}\right)=266 \times 10^{-3} \mathrm{~cm}^{-1}}

Hence, the answer is 266

Posted by

vishal kumar

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