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Integer Answer Type Questions: Electrochemistry | JEE Advanced - Notes | Study Chemistry 35 Years JEE Main & Advanced Past year Papers - JEE
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Integer Value Correct Type
1. All the energy released from the reaction X → Y, ΔrG° = –193 kJ mol–1 is used for oxidizing M+ as M+ → M3+ + 2e–, E° = –0.25 V
Under standard conditions, the number of moles of M+ oxidized when one mole of X is converted to Y is [F = 96500 C mol–1] (JEE Adv. 2015)
Ans: 4
Solution :
Hence DG° for oxidation will be
ΔG° = – nFE°
= –2 × 96500 × (–0.25) = 48250 J = 48.25 kJ
48.25 kJ energy oxidises one mole M+
∴ 193 kJ energy oxidises 
mole M+ = 4 mole M+
2. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If 
the difference in their pKa values, pKa(HX) – pKa (HY), is (consider degree of ionization of both acids to be <<1) (JEE Adv. 2015)
Ans: 3
Solution :
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