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Question

# 1.0 g magnesium atoms in vapour phase absorbs 50.0 kJ of energy to convert all Mg into Mg ions. The energy absorbed is needed for the following changes : Mg(g)⟶Mg+(g)+e;ΔH=740kJmol−1 Mg+(g)⟶Mg2+(g)+e;ΔH=1450kJmol−1Find out the % of Mg+ and Mg2+ in the final mixture.

A
% Mg+=68.28% , % ofMg2+=31.72%
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B
% Mg+=58.28% , % ofMg2+=41.72%
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C
% Mg+=78.28% , % ofMg2+=21.72%
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D
None of these
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Solution

## The correct option is A % Mg+=68.28% , % ofMg2+=31.72%MolesofMg=124Let a mole of Mg+ and b mole of Mg2+ are present, then a+b=124 ... (i)Also, 50=a×740+b×2190 ... (ii)∴ From eqs. (i) and (ii), a=2.845×10−2 b=1.322×10−2 % ofMg+=68.28% % ofMg2+=31.72%

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