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),
\(\Lambda_m = \kappa V\)
Calculate \(\Lambda_m\) for aluminium sulphate. When \(\Lambda_m\) for aluminium chloride, sulphuric acid and hydrochloric acid are 236.2, 459.8 and 121.4 S cm\(^2\) mol\(^{-1}\) respectively:"
- A \(574.6 S cm ^2 mol^{-1}\)
- B \(907.0 S cm ^2 mol^{-1}\)
- C \(178.6 S cm ^2 mol^{-1}\)
- D \(1123.4 S cm ^2 mol^{-1}\)
Answer & Solution
Correct Answer
(D) \(1123.4 S cm ^2 mol^{-1}\)
Step-by-step Solution
Detailed explanation
\(\Lambda_m(Al_2(SO_4)_3) = 2\Lambda_m(AlCl_3) + 3\Lambda_m(H_2SO_4) - 6\Lambda_m(HCl)\) \(\Lambda_m(Al_2(SO_4)_3) = 2(236.2) + 3(459.8) - 6(121.4)\) \(\Lambda_m(Al_2(SO_4)_3) = 472.4 + 1379.4 - 728.4\) \(\Lambda_m(Al_2(SO_4)_3) = 1123.4 \text{ S cm}^2 \text{ mol}^{-1}\)
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