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
Options:
A .  31% Mg+ + 69 % Mg+
B .  69% Mg+ + 31 % Mg+
C .  86% Mg+ + 14% Mg+2
D .  14% Mg+ + 86% Mg+2
Answer: Option B : B Energy absorbed for converting Mg(g)→Mg(g)+ = 750 kJ Energy left unconsumed = 1200 = 750 = 450 kJ. This energy will be required to convertMg(g)+ toMg(g)+2 Thus % ofMg(g)+2 = 4501450× 100 = 31% % ofMg(g)+ = 100 - 31 = 69%
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