Standard enthalpy of vapourisation for CCl 4 is 30 . 5 kJ mol - 1 . Heat required for vapourisation of 284 g of CCl 4 at constant temperature is __________ kJ . (Given molar mass in gmol - 1 ; C = 12 , Cl = 35 . 5 ) Round off your answer to the nearest integer.

Step-by-step solution

The enthalpy of vapourisation ∆ H of CCl 4 = 30 . 5 kJ / mol q p = ∆ H = 30 . 5 kJ / mol No. of moles of CCl 4 = given mass molar mass = 284 154 = 1 . 844 mol Hence, the heat required for the vapourisation of 1 . 844 mol of CCl 4 = 1 . 844 × 30 . 5 Heat required = 56 . 242 kJ / mol ≈ 56 kJ / mol