The enthalpy of vapourisation of CCl4

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