Question

A $KCl$ solution of conductivity $0.14S{m}^{-1}$ 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

Answer: 57

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

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

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

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

$=0.5695S{m}^{-1}$

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