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

Question

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: $1 eV = 96 .5 kJ {mol}^{- 1}$ ]

TestHub · Accepted Answer

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

Chemistry - PERIODIC TABLE Question with Solution | TestHub

Doubts & Discussion

JEE Advanced 2028 (Class 11)