Question

Question #8e917

Answer 1

`2 * 10^(23)`

Explanation:

The trick here is to realize that the molar mass of oxygen gas tells you the mass of `6.022 * 10^(23)` molecules of oxygen gas.

Since oxygen gas has a molar mass of about `"32.0 g mol"^(-1)`, you can say that `6.022 * 10^(23)` molecules of oxygen gas will have a mass of `"32.0 g"`.

This is the case because `1` mole of oxygen gas must contain a number of molecules of oxygen gas equal to Avogadro's constant.

`color(blue)(ul(color(black)("1 mole O"_2 = 6.022 * 10^(23)color(white)(.)"molecules O"_2)))`

So if `1` mole of oxygen gas has a mass of `"32.0 g"`, you can say that `6.022 * 10^(23)` molecules of oxygen gas will have mass of `"32.0 g"`.

This means that your sample contains

`8 color(red)(cancel(color(black)("g"))) * (6.022 * 10^(23)color(white)(.)"molecules O"_2)/(32.0color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(2 * 10^(23)color(white)(.)"molecules O"_2)))`

The answer must be rounded to one significant figure, the number of sig figs you have for the mass of oxygen gas.