First and second ionization energies of magnesium are 7.646 and 15.035 eV respectively. The amount of energy in kJ needed to convert all the atoms of magnesium into ${Mg}^{2 +}$ ions present in 12mg of magnesium vapours is : [Given 1e $V = 96 .5 kJ {mol}^{- 1}$ ]

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1.1

1.5

2.0

0.5

answer is A.

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Detailed Solution

${Total IE of Mg (ieIE}_{1} + {IE}_{2}) \begin{array}{l} = 7 .646 + 15 .035 \\ = 22 .681 \times 96 .5 kJ {mol}^{- 1} \end{array} For 24 g \to 22 .681 \times 96 .5 kJ {mol}^{- 1} \begin{array}{l} 12 \times 10^{- 3} g \to ? \\ = 1 .1 \end{array}$

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