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Question

One mole of magnesium in the vapour state absorbed 1200 kJ mol1 energy . If the first and second ionisation energies of Mg are 750 kJ mol1 and 1450 kJ mol1 respectively, the final composition of the mixture is :

A
86% Mg +14% Mg2+
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B
69% Mg++31% Mg2+
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C
14% Mg++86% Mg2+
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D
31% Mg++69% Mg2+
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Solution

The correct option is B 69% Mg++31% Mg2+
Energy absorbed in the ionisation of 1 mole of Mg to Mg+(g) =750 kJ

Energy left unused =1200750 =450 kJ

% of Mg+(g) converted into Mg2+(g)

=4501450×100=31%

Hence, the % of Mg+=10031=69%

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