How much energy is required to vaporize 98.6 g of ethanol (C₂H₅OH) at its boiling point if its Δ_vapH is 40.5 kJ mol^-1?

Name: How much energy is required to vaporize 98.6 g of ethanol (C 2 H 5 OH) at its boiling point if its Δ vap H is 40.5 kJ mol -1 ?
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Question

Quizplus · Accepted Answer

To find the energy required, first calculate the moles of ethanol: 98.6 g / 46.08 g/mol = 2.14 mol. Then multiply by the enthalpy of vaporization: 2.14 mol * 40.5 kJ/mol = 86.7 kJ.

How Much Energy Is Required to Vaporize 98