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ChemistryInteger+4 / -1Question 71 of 75

JEE Main 2026 · 6 Apr · Shift 1, Chemistry question 71

First and second ionization enthalpies of lithium are 520 kJ mol−1^{-1} and 7297 kJ mol−1^{-1} respectively. Energy required to convert 3.5 mg lithium (g) into LiX2+\ce{Li^2+}(g) [Li(g)→LiX2+(g)\ce{Li(g) -> Li^2+(g)}] is ________ kJ mol−1^{-1}. (nearest integer)

[Molar mass of Li = 7 g mol−1^{-1}]

Answer and worked solution

Official answer: 4

Moles. 3.5 mg of Li is 3.5×10−37=5×10−4\tfrac{3.5\times10^{-3}}7=5\times10^{-4} mol.

Energy.
(520+7297)×5×10−4=3.91 kJ≈4.(520+7297)\times5\times10^{-4}=3.91\text{ kJ}\approx4.

Answer: 4

Topics:Ionisation enthalpyMole concept
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