Question

Question #1550e

Answer 1

Here's how you can do that.

Explanation:

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.