Question #8e917

1 Answer
Jan 4, 2018

Answer:

#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.