Question

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).

Answer 1

$\kappa=\frac{1}{R} \cdot G^{*}$

For same conductivity cell, $G^{*}$ is constant and hence $\kappa \cdot R .=$ constant .

$\therefore 0.14 \times 4.19=\kappa \times 1.03$

or, $\kappa$ of $HCl$ solution $=\frac{0.14 \times 4.19}{1.03}$

$=0.5695\, Sm ^{-1}$

$=56.95 \times 10^{-2} Sm ^{-1} \approx 57 \times 10^{-2} Sm ^{-1}$