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Chemistry · 2024 · 3 marks
CBSE 2024 · Region 1 · Set 1 · Q22
When a certain conductivity cell was filled with 0.05 M KCl solution, it has a resistance of 100 ohm at 25°C. When the same cell was filled with 0.02…
When a certain conductivity cell was filled with $\displaystyle 0.05$ M KCl solution, it has a resistance of $\displaystyle 100$ ohm at $\displaystyle 25$°C. When the same cell was filled with $\displaystyle 0.02$ M $\displaystyle \mathrm{AgNO}_{3}$ solution, the resistance was $\displaystyle 90$ ohm. Calculate the conductivity and molar conductivity of $\displaystyle \mathrm{AgNO}_{3}$ solution. $\displaystyle 3$ (Given : Conductivity of $\displaystyle 0 \cdot 05 \mathrm{M} \mathrm{KCl}$ solution $\displaystyle =1 \cdot 35 \times 10^{-2} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$ )
Marking-scheme solution
Cell constant \(\displaystyle \left(\mathrm{G}^{*}\right)=\) Conductivity × Resistance
\[\begin{aligned}
& =1.35 \times 10^{-2} \times 100 \\
& =1.35 \mathrm{~cm}^{-1}
\end{aligned}
\]
Cell constant \(\displaystyle \left(\mathrm{G}^{*}\right)=\) Conductivity × Resistance
\[\begin{aligned}
& 1.35 \mathrm{~cm}^{-1}=\mathrm{k} \times 90 \\
& 1.35 / 90=\mathrm{k} \\
& \mathrm{k}=0.015 \mathrm{Scm}^{-1}
\end{aligned}
\]
Molar conductivity \(\displaystyle \left(\mathbf{\Lambda}_{\mathrm{m}}\right)=\mathrm{k} \times 1000 / \mathrm{C}\)
\[\begin{aligned}
& =\frac{0.015 \times 1000}{0.02} \\
& =750 \mathrm{Scm}^{2} / \mathrm{mol}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.