The heat of vaporization of benzene, C6H6, is 30.7 kJ/mol at its boiling point of 80.1 °C. How much energy in the form of heat is required to vaporize 134 g benzene at its boiling point?

Answer 1

The amount of energy required to vaporize 134 g of benzene at its boiling point is 4,072 kJ. This can be calculated using the following equation:

Energy = (Mass of benzene (g) * Heat of vaporization (kJ/mol)) / Molecular weight of benzene (g/mol)

Energy = (134 g * 30.7 kJ/mol) / 78.11 g/mol

Energy = 4,072 kJ