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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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62% Mg²⁺ and 38% Mg⁺

59% Mg²⁺ and 41% Mg⁺

41% Mg²⁺ and 59% Mg⁺

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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