When 92.0 g of ethanol (C2H5OH) is vaporized at its boiling point of 78.3°C, 78.6 kJ of energy is required. What is the approximate molar heat of vaporization of ethanol in kJ/mol?
Answer 1
(78.6 kJ) / (92.0 g / (46.0684 g C2H5OH/mol)) = 39.4 kJ/mol
So answer G.