NTA Abhyas JEE Main2020ChemistryClassification of Elements and Periodicity in PropertiesPractice
First and second ionization energies of magnesium are 7.646 and 15 .035 eV respectively. The amount of energy in kJ/mol needed to convert all the atoms of Magnesium into Mg 2+ ions present in 12 mg of magnesium vapours is: (Report your answer by multiplying with 10 and round it upto nearest integer) [Given: 1 eV = 96 .5 kJ mol − 1 ]
Correct answer
11
Step-by-step solution
12 mg 24 g = 0.5 × 10 − 3 moles Energy per atom = 7 .646 + 15 .035 = 22 .68 eV 22 .68 × 96 .48 = 21 .88 × 10 2 kJ/mol E (needed) 2 1 . 8 8 × 1 0 2 × 0 . 5 × 1 0 - 3 10 .94   ×   10 − 1   =   1 .094   kJ/mol