A KCl solution of conductivity 0.14 , S m^{-1 | vedclass

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(D) The conductivity $\kappa$ is related to resistance $R$ and cell constant $G^{*}$ by the formula: $\kappa = \frac{1}{R} \cdot G^{*}$For the same co

Vedclass · Accepted Answer

$57$

Vedclass · Answer

(D) The conductivity $\kappa$ is related to resistance $R$ and cell constant $G^{*}$ by the formula: $\kappa = \frac{1}{R} \cdot G^{*}$For the same conductivity cell,$G^{*}$ is constant,therefore $\kappa \cdot R = G^{*} = 	ext{constant}$.For the $KCl$ solution: $\kappa_{KCl} \cdot R_{KCl} = 0.14 	imes 4.19 = 0.5866 \, S$.For the $HCl$ solution: $\kappa_{HCl} \cdot R_{HCl} = 0.5866 \, S$.$\kappa_{HCl} = \frac{0.5866}{1.03} \approx 0.5695 \, S m^{-1}$.Converting to the required units: $0.5695 \, S m^{-1} = 56.95 	imes 10^{-2} \, S m^{-1}$.Rounding to the nearest integer,we get $57 	imes 10^{-2} \, S m^{-1}$.

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