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JEE CHEMISTRY MCQ: CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

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Application +4 | -1

The first, second and third ionisation energies of

{Al}

are 578,1817 and

2745 \mathrm{~kJ} \mathrm{~mol}^{-1}

respectively. Calculate the energy required to convert all the atoms of

{Al}

{Al}^{+3}

present in

270{mg}

{Al}

vapours

View answer and explanation

Correct option: B

51.40 \mathrm{~kJ}

The correct answer is 51.40 kJ because the energy required to convert all Al atoms to Al+3 in 270 mg of Al vapor is 51.40 kJ, calculated by summing the ionization energies and multiplying by the number of moles of Al.

JEE CHEMISTRY MCQ: CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

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