Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper

JEE : Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

The document Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev is a part of the JEE Course Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper.
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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 :

Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

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 Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev 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  Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev  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 :

Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

 

 

 

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