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Question

# 1 mole of magnesium in the vapour state absorbed 1200 kJ molâˆ’1 of energy. If the first and second ionization energies of Mg are 750 and 1450 kJ molâˆ’1 respectively, the final composition of the mixture is

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

## The correct option is B 69 % Mg+ + 31% Mg2+Energy absorbed in the ionization of 1 mole of Mg (g) to Mg+ (g) = 750 kJ. Energy left unconsumed = 1200 - 750 = 450 kJ This energy is used to convert Mg+ (g) to Mg2+(g), but we actually need to 1450 kJ to convert it entirely. So, instead of all molecules absorbing a little bit of energy, a certain fraction of them only would absorb enough energy and get ionized while the rest remain in the same state. Remember, energy is quantized. Thus, % ofMg2+(g) = 450 x 100/ 1450 = 31 % % ofMg+ (g) = 100 - 31 = 69%

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