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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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$41 % M g^{2 +} a n d 59 % M g^{+}$

$59 % M g^{2 +} a n d 41 % M g^{+}$

$62 % M g^{2 +} a n d 38 % M g^{+}$

$38 % M g^{2 +} a n d 62 % M g^{+}$

answer is B.

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Detailed Solution

$I P_{1} = 750$ , Energy available for $I P_{2} = 1300 - 750 = 550$
1450kj will convert $M g^{+} (g) t o M g^{2 +} (g)$ = 1 mole
550kj will convert $M g^{+} (g) t o M g^{2 +} (g)$ = 0.38 mole

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