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Question

# First and second ionisation energy of Mg(g) are 720 k/J mol and 1440 kJ/mol respectively. Calculate the % of Mg+ ions if one gram of Mg(s) absorbs 50 kJ of energy. (Atomic mass of Mg is 24 amu.)

A
31.65%
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B
21.65%
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C
11.65%
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D
41.65%
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Solution

## The correct option is A 31.65%Amount of Mg atoms=124=4.167×10−2molEnergy absorbed in ionizing Mg to Mg+=4.167×10−2×740=30.84KJEnergy absorbed in ionizing Mg+ to Mg2+=(50−30.84)KJ=19.16KJAmount of Mg+ converted to Mg2+=19.161450=1.321×10−2molAmount of Mg+ remaining as such=4.167×10−2−1.321×10−2=2.846×10−2molComposition of final mixture would be as follows:% of Mg+=2.846×10−24.167×10−2×100=68.3%% of Mg2+=100−68.3=31.7 %

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