Q.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).
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Start your 14-day free trial to unlock the full solution →The heat required at constant pressure equals the enthalpy change for the process. Using the given enthalpy of vaporisation and the number of moles in 284 g of CCl₄, the answer is 56.2 kJ.
The key idea here is that enthalpy of vaporisation () is the heat absorbed when one mole of a substance vaporises at constant pressure. Since the problem asks for heat at constant pressure, that heat is the enthalpy change — no extra conversion needed. The only work is to find how many moles are in 284 g and multiply.
- Find the number of moles. Molar mass of CCl₄ is given as 154 g mol⁻¹.
Simplify:
(You can keep it as a fraction for exactness — we’ll use the decimal for clarity.)
- Apply the enthalpy of vaporisation. means each mole requires 30.5 kJ of heat at constant pressure. So total heat is:
Compute:
Rounding to three significant figures (since 30.5 has three and 284 has three): …
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