Question

The first second and third ionisation energies of Al
are 578, 1817 and 2745 kJ mol- respectively.
Calculate the energy required to convert all the atoms
of Al to AI*3 present in 270 mg of Al vapours
(1) 5140 kJ
(2) 51.40 kJ
(3) 2745 kJ
(4) 514.0 kJ​

Answer 1

Answer:

add all three energies and then divide by 100 since 0.01 mole of Al are there

therefore the answer is 51.40 kJ