CBSE 2024 · Region 5 · Set 2 · Q18 · 2 marks
Resistance of a conductivity cell filled with $\displaystyle 0.2 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{KCl}$ solution is $\displaystyle 200 \Omega$. If the resistance of the same cell when filled with $\displaystyle 0.05 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{KCl}$ solution is $\displaystyle 620 \Omega$, calculate the conductivity and molar conductivity of $\displaystyle 0.05 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{KC} l$ solution. The conductivity of $\displaystyle 0.2 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{KC} l$ solution is $\displaystyle 0.0248 \mathrm{~S} \mathrm{~cm}^{-1}$.
Marking-scheme solution
Cell constant \(\displaystyle =G^{*}=\) conductivity × resistance \(\displaystyle =0.0248 \mathrm{~S} / \mathrm{cm} \times 200 \mathrm{ohm}=4.96 \mathrm{~cm}^{-1}\) Conductivity of \(\displaystyle 0.05 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{KCl}\) solution = cell constant / resistance
\[=* G / R=4.96 / 620=0.008 \mathrm{~S} \mathrm{~cm}^{-1}
\] Molar conductivity \(\displaystyle =\mathbf{\Lambda}_{\mathrm{m}}=\frac{k \times 1000}{c} =0.008 \times 1000 / 0.05=160 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
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CBSE Class 12 Chemistry past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.