# Question 35f15

Apr 2, 2016

$\Delta {H}_{\text{vap" = "8.85 kJ mol}}^{- 1}$

#### Explanation:

The molar enthalpy of vaporization, $\Delta {H}_{\text{vap}}$, tells you how much heat must be absorbed by one mole of a given substance in order for a liquid to vapor phase change to take place.

In your case, you know that $3.21$ moles of a given substance require $\text{28.4 kJ}$ of heat in order to go from liquid at its boiling point to vapor at its boiling point.

Your goal here is to determine how much heat is required in order for one mole of this substance to undergo this phase change, so you can say that

1color(red)(cancel(color(black)("mole"))) * "28.4 kJ"/(3.21color(red)(cancel(color(black)("mole2")))) = "8.847 kJ"

So, if one mole of this substance requires $\text{8.847 kJ}$ of heat in order to go from liquid to vapor, it follows that the substance's molar enthalpy of vaporization will be equal to

DeltaH_"vap" = color(green)(|bar(ul(color(white)(a/a)"8.85 kJ mol"^(-1)color(white)(a/a)|)))#

The answer is rounded to three sig figs.