Solve this following


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


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

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

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

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

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

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


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