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Question

The enthalpy of vaporization for chlorine is 20.41kJ/mol. How much heat is required to vaporize 223.05g of chlorine liquid at its boiling point? Report your answer in units of kJ. ( show steps)

Answer 1

The enthalpy of vaporization is the amount of heat needed to vaporize 1 mol of a substance at it's boiling point. In this case for chlorine (Cl₂), it is 20.41 kJ per mole. So, if we have 223.05 g of Cl₂, how much heat do we need to vaporize it? Well, we first have to find out how many moles of Cl₂ we have, and then multiply that by 20.41 kJ per mole.

molar mass Cl₂ = 70.91 g / mol

moles Cl₂ present = 223.05 g x 1 mol / 70.91 g = 3.146 moles

Heat needed to vaporize this = 3.146 moles x 20.41 kJ / mole = 64.20 kJ