The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 $\Omega$. If the conductivity of 0.01 M KCl solution at 298 K is 0.152 $\times$ 10$-$3 S cm$-$1, then the cell constant of the conductivity cell is ____________ $\times$ 10$-$3 cm$-$1.
Answer (integer)
266
Solution
Molarity of $\mathrm{KCl}$ solution $=0.1 ~\mathrm{M}$
<br/><br/>
$$
\begin{array}{ll}
\text { Resistance } & =1750 ~\mathrm{ohm} \\\\
\text { Conductivity } & =0.152 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1} \\\\
\text { Conductivity } & =\frac{\text { Cell constant }}{\text { Resistance }} \\\\
\therefore \text { Cell constant } & =0.152 \times 10^{-3} \times 1750 \\\\
& =266 \times 10^{-3} \mathrm{~cm}^{-1}
\end{array}
$$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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