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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

A

31% Mg++69% Mg+2

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B

69% Mg++31% Mg+2

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C

86% Mg++14% Mg+2

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D

14% Mg++86% Mg+2

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Solution

## The correct option is B 69% Mg++31% Mg+2 Energy absorbed for converting Mg(g) → Mg(g)+ = 750 kJ Energy left unconsumed = 1200 = 750 = 450 kJ. This energy will be required to convert Mg(g)+ to Mg(g)+2 Thus % of Mg(g)+2 = 4501450 × 100 = 31% % of Mg(g)+ = 100 - 31 = 69%

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