Question

The first, second and third ionisation energies of Al
are 578, 1817 and 2745 kJ mol-7 respectively.
Calculate the energy required to convert all the atoms
of Al to Al+3 present in 270 mg of Al vapours
(1) 5140 kJ
(2) 51.40 kj
(3) 2745 kJ
(4) 514.0 kj​

Answer 1

B) 51.4 KJ is the correct answer....

Answer 2

270 mg=.27g

moles of aluminum in 270 mg =.27/27 =.01moles

first ionization energy for.01 moles=5.78

second ionisation energy for .01 moles=18.17

third ionisation energy for.01moles=27.45

so total energy required=18.17+27.45+5.78

=51.4

hope it helps you