Question

Question #70282

Answer 1

`"82.3 kJ"`

Explanation:

In order to be able to answer this question, you must know the enthalpy of vaporization, `DeltaH_"vap"`, for liquid oxygen, which you'll find listed as

`DeltaH_"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^@"C"`, to vapor at its boiling point.

In this case, `1` mole of liquid oxygen requires `"3.41 kJ"` of heat in order to perform that liquid `->` 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, `"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.