JEE MAIN - Chemistry (2022 - 24th June Evening Shift - No. 17)
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
266
Explanation
Molarity of $\mathrm{KCl}$ solution $=0.1 ~\mathrm{M}$
$$ \begin{array}{ll} \text { Resistance } & =1750 ~\mathrm{ohm} \\\\ \text { Conductivity } & =0.152 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1} \\\\ \text { Conductivity } & =\frac{\text { Cell constant }}{\text { Resistance }} \\\\ \therefore \text { Cell constant } & =0.152 \times 10^{-3} \times 1750 \\\\ & =266 \times 10^{-3} \mathrm{~cm}^{-1} \end{array} $$
$$ \begin{array}{ll} \text { Resistance } & =1750 ~\mathrm{ohm} \\\\ \text { Conductivity } & =0.152 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1} \\\\ \text { Conductivity } & =\frac{\text { Cell constant }}{\text { Resistance }} \\\\ \therefore \text { Cell constant } & =0.152 \times 10^{-3} \times 1750 \\\\ & =266 \times 10^{-3} \mathrm{~cm}^{-1} \end{array} $$
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