Please help me with chemistry math problems: URGENT PLEASE ANSWER

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

Answer 1

Hello, Maddie,

The heat of vaporization is given in kJ/mole. Once we convert 90.8 grams of ethanol to moles ethanol, we can multiply by the heat of vaaporization to find the heat necessary to vaporize the ethanol. Since the question dioes not say that a temperature change occured, we'll assume that the ethanol is already at it's boiling point, and the resulting gas stays at the same temperature (otherwise, we'd need to calculate the heats involved with temperature changes).

Ethanol, C₂H₅OH, is 46.07 grams/mole. We have 90.8 grams/(46.07 grams/mole) = 1.97 moles C₂H₅OH

(1.97 moles C₂H₅OH)*(39.3 kJ/(mole C₂H₅OH)) = 77.45 kJ (44.5 kJ with 3 sig figs)

Bob