# Question 1550e

Jan 29, 2018

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 $\text{0.00338 M}$ of chlorine gas in a container that has a volume of $\text{20.0 L}$.

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 $\text{1 L}$, tells you that this sample contains $0.00338$ moles of chlorine gas for every $\text{1-L}$ 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.

$0.0676 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Cl"_2))) * "70.906 g"/(1color(red)(cancel(color(black)("mols Cl"_2)))) = color(darkgreen)(ul(color(black)("4.79 g}}}}$

The answer is rounded to three sig figs.

To find the number of atoms of chlorine, $\text{Cl}$, present in this sample of chlorine gas, ${\text{Cl}}_{2}$, use the fact that $1$ mole of chlorine gas contains $6.022 \cdot {10}^{23}$ molecules of chlorine gas $\to$ 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

$4.071 \cdot {10}^{22} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{molecules Cl"_2))) * "2 atoms Cl"/(1 color(red)(cancel(color(black)("molecule Cl"_2)))) = color(darkgreen)(ul(color(black)(8.14 * 10^(22) quad "atoms Cl}}}}$

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