Question
One mole of magnesium in the vapour state absorbed 1200 kJ mole−1 of energy. If the first and second ionization energies of Mg are 750 and 1450 kJ mole−1 respectively, the final composition of the mixture is
Answer: Option B
:
B
Energy absorbed for converting Mg(g) → Mg(g)+ = 750 kJ
Energy left unconsumed = 1200 = 750 = 450 kJ.
This energy will be required to convertMg(g)+ toMg(g)+2
Thus % ofMg(g)+2 = 4501450 × 100 = 31%
% ofMg(g)+ = 100 - 31 = 69%
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:
B
Energy absorbed for converting Mg(g) → Mg(g)+ = 750 kJ
Energy left unconsumed = 1200 = 750 = 450 kJ.
This energy will be required to convertMg(g)+ toMg(g)+2
Thus % ofMg(g)+2 = 4501450 × 100 = 31%
% ofMg(g)+ = 100 - 31 = 69%
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