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  • A KCl solution of conductivity shows a resistance of in a conductivity cell. If the same cell is filled

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

Answers (1)

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We know the relation between conductivity and resistance

\kappa=\frac{1}{R} \cdot G^{*}\mathrm{\kappa=\frac{1}{R} \cdot G^{*}=\frac{1}{R} \times\left(\frac{l}{A}\right)}

Where 

\begin{aligned} &\kappa=\text { Conductivity } \\ &\frac{l}{A}=G^*=\text { Cell Constant } \\ &\mathrm{R}=\text { Resistance } \end{aligned}

For the same conductivity cell, Cell constant will be constant

\text { hence } \kappa . \mathrm{R} .=\text { constant }

\mathrm{\therefore \kappa_{KCl} \times R_{KCl}=\kappa_{HCl} \times R_{HCl}}

0.14 \times 4.19=\kappa \times 1.03

\kappa \text { of } \mathrm{HCl} \text { solution }=0.5695 \mathrm{Sm}^{-1}

\kappa \text { of } \mathrm{HCl} \text { solution }=56.95 \times 10^{-2} \mathrm{Sm}^{-1}

\kappa \text { of } \mathrm{HCl} \text { solution }\approx 57 \times 10^{-2} \mathrm{Sm}^{-1}

Ans = 57

Posted by

Kuldeep Maurya

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