Question #1550e

Every time a problem gives you the molarity of a solution and its volume, it's actually giving you the number of moles of solute present in the sample.

In your case, you know that you have a concentration of #"0.00338 M"# of chlorine gas in a container that has a volume of #"20.0 L"#.

The molarity of the gas, which is defined as the number of moles of solute, in your case chlorine gas, present in a volume of #"1 L"#, tells you that this sample contains #0.00338# moles of chlorine gas for every #"1-L"# volume.

You can thus say that your container has a total of

#20.0 color(red)(cancel(color(black)("L"))) * overbrace("0.00338 moles Cl"_2/(1color(red)(cancel(color(black)("L")))))^(color(blue)(="0.00338 M Cl"_2)) = "0.0676 moles Cl"_2#

To convert the number of moles to grams, you can use the molar mass of chlorine gas.

#0.0676 color(red)(cancel(color(black)("moles Cl"_2))) * "70.906 g"/(1color(red)(cancel(color(black)("mols Cl"_2)))) = color(darkgreen)(ul(color(black)("4.79 g")))#

The answer is rounded to three sig figs.

To find the number of atoms of chlorine, #"Cl"#, present in this sample of chlorine gas, #"Cl"_2#, use the fact that #1# mole of chlorine gas contains #6.022 * 10^(23)# molecules of chlorine gas #-># think Avogadros constant here.

This means that your sample contains

#0.0676 color(red)(cancel(color(black)("moles Cl"_2))) * (6.022 * 10^(23) quad "molecules Cl"_2)/(1color(red)(cancel(color(black)("mole Cl"_2)))) = 4.071 * 10^(22) quad"molecules Cl"_2#

Finally, you know that every molecule of chlorine gas contains #2# atoms of chlorine, which means that your sample will contain

#4.071 * 10^(22) color(red)(cancel(color(black)("molecules Cl"_2))) * "2 atoms Cl"/(1 color(red)(cancel(color(black)("molecule Cl"_2)))) = color(darkgreen)(ul(color(black)(8.14 * 10^(22) quad "atoms Cl")))#

Once again, the answer is rounded to three sig figs.