# Question #067de

##### 1 Answer

#### Explanation:

All you have to do here is to use the **density** of liquid oxygen given to you to figure out the *mass* of the sample.

You know that liquid oxygen at its boiling point has a density of

This means that your sample will have a mass of

#3.0 color(red)(cancel(color(black)("L"))) * "1.14 kg"/(1color(red)(cancel(color(black)("L")))) = "3.42 kg"#

Now, the **heat of vaporization** tells you the amount of heat needed in order to boil **unit of mass** of a given substance at its boiling point.

In your case, liquid oxygen is said to have a heat of vaporization equal to

You can thus say that your sample will require

#3.42 color(red)(cancel(color(black)("kg"))) * "213 kJ"/(1color(red)(cancel(color(black)("kg")))) = "728.5 kJ"#

Rounded to two **sig figs**, the number of sig figs you have for the volume of the sample, and expressed in *joules*, the answer will be--keep in mind that you have

#728.5 color(red)(cancel(color(black)("kJ"))) * (10^3color(white)(.)"J")/(1color(red)(cancel(color(black)("kJ")))) = color(darkgreen)(ul(color(black)("730,000 J")))#