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JEE Mains · Chemistry · STD 11 - 3. Classification of elements and periodicity in properties

First and second ionization enthalpies of lithium are \(520\) kJ mol\(^{-1}\) and \(7297\) kJ mol\(^{-1}\) respectively. Energy required to convert \(3.5\) mg lithium (g) into \(\text{Li}^{2+}\)(g) \([\text{Li(g)} \rightarrow \text{Li}^{2+}\text{(g)}]\) is _______ kJ mol\(^{-1}\). (nearest integer) [Molar mass of Li \(= 7\) g mol\(^{-1}\)]

  1. A 3905
  2. B 3906
  3. C 3907
  4. D 3909
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Answer & Solution

Correct Answer

(D) 3909

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Detailed explanation

The total energy required to convert \(1\) mole of \(\text{Li(g)}\) to \(\text{Li}^{2+}\text{(g)}\) is the sum of the first and second ionization enthalpies: \(E = IE_1 + IE_2 = 520 + 7297 = 7817 \text{ kJ mol}^{-1}\) Given mass of lithium…
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