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)

Solution

Chemistry ma’am · 3 steps

Turn the grams into moles first, then multiply by the enthalpy per mole.

1. Molar mass. CCl₄ has one C and four Cl, so M = 12 + 4 × 35.5 = 154 g mol⁻¹.
2. Moles present. n = 284/154 = 1.844 mol.

   q = n × ΔᵥₐₚH° = 284/154 × 30.5 kJ = 56.25 kJ

3. So. The heat needed is about 56.25 kJ, which rounds to 56.

The common slip: 61. Taking 142 g mol⁻¹ as the molar mass counts only the four chlorines and leaves out carbon. That gives 2 mol and 61 kJ, not 56.

The idea to keep. Enthalpy of vapourisation is given per mole. Find the moles from the mass and the molar mass, then multiply. Add up every atom when you work out the molar mass.

Answer from the NTA final answer key. Solution by PrepWiser.