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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=740kJmol1
Mg+(g)Mg2+(g)+e;ΔH=1450kJmol1

Find 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=124

Let 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=2.845×102

b=1.322×102

% ofMg+=68.28%

% ofMg2+=31.72%

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