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Standard X

Chemistry

Ionization Enthalpy

One mole of m...

Question

One mole of magnesium in the vapour state absorbed 1200 kJ ${m o l e}^{- 1}$ of energy. If the first and second ionization energies of Mg are 750 and 1450 kJ ${m o l e}^{- 1}$ respectively, the final composition of the mixture is

31% Mg⁺+69% Mg⁺²

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69% Mg⁺+31% Mg⁺²

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86% Mg⁺+14% Mg⁺²

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14% Mg⁺+86% Mg⁺²

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Solution

The correct option is B
69% Mg⁺+31% Mg⁺²

Energy absorbed for converting $M g_{(g)}$ $\to$ ${M g_{(g)}}^{+}$ = 750 kJ

Energy left unconsumed = 1200 = 750 = 450 kJ.

This energy will be required to convert ${M g_{(g)}}^{+}$ to ${M g_{(g)}}^{+ 2}$

Thus % of ${M g_{(g)}}^{+ 2}$ = $\frac{450}{1450}$ $\times$ 100 = 31%

% of ${M g_{(g)}}^{+}$ = 100 - 31 = 69%

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Similar questions

Q. 1 mole of magnesium in the vapour state absorbed 1200 kJ

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One mole of magnesium in the vapour state absorbed 1200 kJ ${m o l e}^{- 1}$ of energy. If the first and second ionization energies of Mg are 750 and 1450 kJ ${m o l e}^{- 1}$ respectively, the final composition of the mixture is

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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

The correct option is B 69% Mg++31% Mg+2 Energy absorbed for converting Mg(g) → Mg(g)+ = 750 kJ Energy left unconsumed = 1200 = 750 = 450 kJ. This energy will be required to convert Mg(g)+ to Mg(g)+2 Thus % of Mg(g)+2 = 4501450 × 100 = 31% % of Mg(g)+ = 100 - 31 = 69%

One mole of magnesium in the vapour state absorbed 1200 kJ ${m o l e}^{- 1}$ of energy. If the first and second ionization energies of Mg are 750 and 1450 kJ ${m o l e}^{- 1}$ respectively, the final composition of the mixture is