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Question

IE1 and IE2 of magnesium are 7.6 eV and 15.0 eV respectively. The amount of energy in kJ needed to convert all the atoms of magnesium into Mg2+ ions present in 36 mg of magnesium vapour will be:
[Given: 1 eV=96.5 kJ mol1].

A
3.27
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B
1.62
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C
2.18
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D
2.51
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Solution

The correct option is A 3.27
Given: MgMg+;IE1=7.6 eV
Mg+Mg2+;IE2=15.0 eV
Hence,
MgMg2+;IE=22.6 eV/atom
Given:
1 eV=96.5 kJmol1

IE in kJ/mol=22.6×96.5=2181 kJmol1

Moles of Mg in 36 mg of Mg vapours,
=36×103 g24 g=1.5×103 mol

1 mol of Mg has an ionisation energy =2181 kJ
1.5×103 mol will have an ionisation energy =1.5×103×2181 kJ
=3271.5×103 kJ

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