CBSE 2022 · Region 2 · Set 3 · Q2 · 2 marks
The resistance and conductivity of a conductivity cell containing $\displaystyle 0.001$ M KCl solution at $\displaystyle 298$ K are $\displaystyle 1200 \Omega$ and $\displaystyle 1.5 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its cell constant and molar conductivity.
Marking-scheme solution
\[\kappa=\frac{1}{R}\left(\frac{l}{A}\right)
\]
\(\displaystyle \implies 1.5 \times 10^{-4} \mathrm{Scm}^{-1}=\frac{1}{1200} \Omega\left(\frac{l}{A}\right)\)
\(\displaystyle \left(\frac{l}{A}\right)=1 \cdot 5 \times 10^{-4} \times 1200 \mathrm{~cm}^{-1}\)
\(\displaystyle \left(\frac{l}{A}\right)=0 \cdot 18 \mathrm{~cm}^{-1}\)
\(\displaystyle \mathbf{\Lambda}_{m}=\frac{\kappa}{C} \times 1000 \Omega^{-1} \boldsymbol{c m}^{2} \boldsymbol{m o l}^{-1}\)
\(\displaystyle =\frac{1 \cdot 5 \times 10^{-4}}{0 \cdot 001} \times 1000 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
\(\displaystyle \mathbf{\Lambda}_{m}=150 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) or \(\displaystyle 150 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
ElectrochemistryConductance of Electrolytic SolutionsApplyvery_short_answereasy
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CBSE Class 12 Chemistry past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.