Question 70282

Jul 11, 2016

$\text{82.3 kJ}$

Explanation:

In order to be able to answer this question, you must know the enthalpy of vaporization, $\Delta {H}_{\text{vap}}$, for liquid oxygen, which you'll find listed as

$\Delta {H}_{\text{vap" = "3.41 kJ mol}}^{- 1}$

https://en.wikipedia.org/wiki/Enthalpy_of_vaporization

Now, the enthalpy of vaporization tells you how much heat is needed in order to convert $1$ mole of liquid oxygen from liquid at its boiling point, i.e. at $- {183}^{\circ} \text{C}$, to vapor at its boiling point.

In this case, $1$ mole of liquid oxygen requires $\text{3.41 kJ}$ of heat in order to perform that liquid $\to$ solid phase change.

The first thing to do here is convert your sample from grams to moles by using oxygen's molar mass. Do not forget that oxygen is a diatomic molecule, ${\text{O}}_{2}$, when using its molar mass!

772 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0 color(red)(cancel(color(black)("g")))) = "24.125 moles O"_2

Now use the enthalpy of vaporization as a conversion factor to determine how much heat is needed to vaporize this many moles of liquid oxygen

24.125 color(red)(cancel(color(black)("moles O"_ 2))) * overbrace("3.41 kJ"/(1color(red)(cancel(color(black)("mole O"_ 2)))))^(color(blue)(=DeltaH_ "vap")) = color(green)(|bar(ul(color(white)(a/a)color(black)("82.3 kJ")color(white)(a/a)|)))#

The answer is rounded to three sig figs.