ExamBro
ExamBro
AP EAMCET · Chemistry · Classification of Elements and Periodicity in Properties

Calculate the energy required to convert all atoms in \(4.8 \mathrm{~g}\) of \(\mathrm{Mg}\) to \(\mathrm{Mg}^{2+}\) in the vapour state. \(\mathrm{IE}_1\) and \(\mathrm{IE}_2\) of \(\mathrm{Mg}\) are \(740 \mathrm{~kJ} / \mathrm{mol}\) and \(1450 \mathrm{~kJ} / \mathrm{mol}\) respectively.

  1. A \(+740 \mathrm{~kJ} / \mathrm{mol}\)
  2. B \(-740 \mathrm{~kJ} / \mathrm{mol}\)
  3. C \(-1450 \mathrm{~kJ} / \mathrm{mol}\)
  4. D \(+438 \mathrm{~kJ} / \mathrm{mol}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(+438 \mathrm{~kJ} / \mathrm{mol}\)

Step-by-step Solution

Detailed explanation

For 1 mole of \(\left(\mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+}\right)\) \(\mathrm{IE}=\mathbb{E E}_1+\mathrm{IE}_2=(740+1450)=2190 \mathrm{~kJ} / \mathrm{mol}\). Number of mole in \(4.8 \mathrm{~g}\) of \(\mathrm{Mg}=4.8 / 24=0.2 \mathrm{~mol}\) For 1 mole energy required…
Same subject
Explore more questions on app