Question

What is the heat required to vaporize 90.8 g of liquid ethanol, C₂H₅OH, at its boiling point? The ΔH° of vaporization of ethanol is 39.3 kJ/mol. Use a molar mass with at least as many significant figures as the data given.

Answer in kJ units

Answer 1

$Moles =\frac{ 90.8 g }{46.07 g/mol }= 1.971 mol$

$∆H°= 1.971mol × 39.3 kJ/mol = 77.4603kJ$