The enthalpy of vapourisation of CCl₄ is 30.5 kJ mol^–1. Calculate the heat required for the vapourisation of 284 g of CCl₄ at constant pressure. (Molar mass of CCl₄ = 154 g mol^–1).

The enthalpy of vapourisation is given for 1 mole of CCl₄ = 30.5 kJ mol^–1

Hence, for 284 g, it will be = (mole no. × 30.5) kJ

Molar mass of CCl₄ = 154 g mol^–1 that means 154 g = 1 mole.

Therefore , 284 g = (284g/154gmol^-1)= 1.84 mole.

Hence,

the heat required for the vapourisation of 284 g of CCl₄ at constant pressure = (1.84 mol ×30.5 KJ mol^-1) kJ

=56.12 KJ