JEE Main 2024 Chemistry question
Standard enthalpy of vapourisation for CCl₄ is 30.5 kJ mol⁻¹. Heat required for vapourisation of 284 g of CCl₄ at constant temperature is kJ. (Given molar mass in g mol⁻¹; C = 12, Cl = 35.5)
Answer: 56
Solution
Molar mass of CCl₄ = 154 g/mol, so moles = 284/154 = 1.84. Heat required = 1.84 × 30.5 ≈ 56 kJ.