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Q.

One mole of magnesium in the vapour state absorbed 1300kJ of energy. If the first and  second ionization energies of Mg are 750 and 1450 kJ /mol respectively, the final composition of mixture is

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a

62% Mg2+ and 38% Mg+

b

59% Mg2+ and 41% Mg+

c

41% Mg2+ and 59% Mg+

d

38% Mg2+ and 62% Mg+

answer is B.

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Detailed Solution

IP1 =750, Energy available for IP2 = 1300 – 750 = 550
1450kj will convert Mg+ (g) to Mg2+ (g) = 1 mole
550kj will convert Mg+ (g) to Mg2+ (g) = 0.38 mole

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