Test SeriesFree PYQsExamsPricing

JEE Main 2026 Chemistry question

ChemistryClassification of Elements and Periodicity in PropertiesJEE Main 2026

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 Li²⁺(g) [Li(g)→Li²⁺(g)] is (nearest integer) [Molar mass of Li=7 g mol⁻¹]

Answer: 3909

Solution

Energy per mole=520+7297=7817 kJ. Amount of Li=3.5×10⁻³/7=5×10⁻⁴ mol, so energy=7817×5×10⁻⁴=3.9085 kJ≈3909 J.

See the full step-by-step solution

Create a free account to read the complete working, and practise questions like this in a timed test.

Create a free account

Question ID VK-015705