The molar enthalpy of vaporization of the substance is 84.3 kj/mol.
What is molar enthalpy of vaporization?
Molar enthalpy of vaporization is the energy required to vaporize one mole of a liquid at its boiling point.
0.433 moles of the substance absorbs 36.5 kj of energy when it is vaporized.
1 mole of the substance will absorb 36.5 kJ × 1 /0.433mol = 84.3 kj/mol
Therefore, the molar enthalpy of vaporization of the substance is 84.3 kj/mol.
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