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Question

First and second ionization energies of Mg are 7.646 and 15.035 eV respectively. The amount of energy in kJ needed to convert all the atoms of magnesium into Mg2+ ions present in 12 mg of magnesium vapours is:


[Given: 1 eV=96.5 kJ mol1]

A
1.5
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B
2.0
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C
1.1
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D
0.5
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Solution

The correct option is A 1.1
The energy required to convert one Mg atom to Mg2+ ion is the sum of first and second ionization potentials = 7.646+15.035=22.681eV

=96.5×22.681=2188.7kJ/mol
The atomic mass of Mg is 24 g/mol.
12 mg of Mg corresponds to 0.5 mmoles.
The amount of energy required to convert 0.5 mmol of Mg atoms to Mg2+ ions = 0.5×103×2188.7=1.1kJ

Hence, option C is correct.

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