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Question
The enthalpy of vapourisation of \[\ce{CCl4}\] is 30.5 kJ mol–1. Calculate the heat required for the vapourisation of 284 g of \[\ce{CCl4}\] at constant pressure. (Molar mass of \[\ce{CCl4}\] = 154 g mol–1).
Solution
As per the information provided in the question, for one mole of \[\ce{CCl4 (154 g)}\], the heat of vaporisation required is 30.5 kJ/mol.
Hence for the vaporisation of 284 g of \[\ce{CCl4}\], we require:
= `284/154 xx 30.5`
= 56.2 kJ
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