Question

What mass is associated with `1*L` of dioxygen gas at `"STP"`?

Answer 1

Well, `22.4*L` of oxygen gas at `"STP"` has a mass of `32.00*g`.

Explanation:

I am a bit cagy when I quote these values, because there are different standards in each different syllabus......

AS far as I know, `"1 mole"` of Ideal Gas occupies `22.4*L` under conditions of `"STP"`.........And of course, we know that `1*mol` of oxygen gas has a mass of `32.0*g` because we deal with the dioxygen molecule; in fact, most of the elemental gases (certainly the ones with any chemistry) are binuclear under standard conditions; this is something you need to know......

And thus............

`"Mass of dioxygen gas"` `=` `(1*Lxx32.0*g*mol^-1)/(22.4*L*mol^-1)` `~=1.5*g`.

Answer 2

`1.43` `"g O"_2`

(or `1` `"g O"_2` with `1` significant figure)

Explanation:

We're asked to find the mass, in `"g"` of `1` `"L"` of oxygen gas at standard temperature and pressure.

Chemists often use this fact: One mole of an (ideal) gas at standard temperature and pressure occupies a volume of `22.4` `"L"`.

(The definition of standard temperature and pressure varies slightly depending on usage, but you'll be using this fact most of the time.)

With that being said, we can use the conversion factor

`(1color(white)(l)"mol")/(22.4color(white)(l)"L")`

to convert the given `1` `"L"` oxygen to moles:

`1cancel("L O"_2)((1color(white)(l)"mol O"_2)/(22.4cancel("L O"_2))) = color(red)(0.0446` `color(red)("mol O"_2`

Now that we know the number of moles of `"O"_2`, we can calculate the number of grams using its molar mass (`32.00` `"g/mol"`):

`color(red)(0.0446)cancel(color(red)("mol O"_2))((32.00color(white)(l)"g O"_2)/(1cancel("mol O"_2))) = color(blue)(1.43` `color(blue)("g O"_2`

Which, if you wish to round to one significant figure, is simply `colorblue)(1` `color(blue)("g O"_2`. (Very rarely are calculations done with only one significant figure.)