CUET · CHEMISTRY · PYQ PAPER 2023
"Answer the question on basis of passage given below:
\(G = \frac{\kappa A}{l}\) =k(both \(A\) and \(l\) are unity in their appropriate units in m or cm)
Molar conductivity of a solution at a given concentration is the conductance of the volume \(V\) of solution containing one mole of electrolyte kept between two electrodes with area of cross section \(A\) and distance of unit length (\(l\)). Therefore,
\(\Lambda_m = \frac{\kappa A}{l}\)
Since \(l = 1\) and \(A = V\) (volume containing 1 mole of electrolyte),
\(\Lambda_m = \kappa V\)
Molar conductance for 0.2 M AgCl solution showing resistance of 40 \(\Omega\) in a cell having distance between electrodes as 2 cm and surface area of 4 cm\(^2\) will be:"
- A \(125\text{ S cm}^2\text{ mol}^{-1}\)
- B \(250\text{ S cm}^2\text{ mol}^{-1}\)
- C \(62.5\text{ S cm}^2\text{ mol}^{-1}\)
- D \(31.25\text{ S cm}^2\text{ mol}^{-1}\)
Answer & Solution
Correct Answer
(C) \(62.5\text{ S cm}^2\text{ mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\( \kappa = \frac{1}{R} \times \frac{l}{A} = \frac{1}{40 \, \Omega} \times \frac{2 \, \text{cm}}{4 \, \text{cm}^2} = 0.0125 \, \text{S cm}^{-1} \)…
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