# #"CHClF"_2# with a mass of #7.21 xx 10^(-2)# kg escaped into a room with a volume of #"200 m"^3#. Assuming #"CHClF"_2# is evenly distributed throughout the air in the room, how do you calculate the number of #"CHClF"_2# molecules in a #"250 cm"^3# sample?

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The Avogadro constant is #6.02 xx 20^23 "mol"^-1#

The Avogadro constant is

##### 1 Answer

#### Explanation:

Your strategy here will be to use the **molar mass** of chlorodifluoromethane, **Avogadro's number** to calculate how many **molecules** of you have in that

Chlorodifluoromethane has a molar mass of *kilograms* to *grams*

#7.21 * 10^(-2) color(red)(cancel(color(black)("kg"))) * (10^3 "g")/(1color(red)(cancel(color(black)("kg")))) = "72.1 g"#

Your sample will thus contain

#72.1 color(red)(cancel(color(black)("g"))) * "1 mole CHClF"_2/(86.47color(red)(cancel(color(black)("g")))) = "0.8338 moles CHClF"_2#

Now that you know how many *moles* of chlorodifluoromethane you have in your sample, use Avogadro's number to convert them to **number of molecules**

#0.8338 color(red)(cancel(color(black)("moles CHClF"_2))) * (6.022 * 10^(23)"molec.")/(1color(red)(cancel(color(black)("mole CHClF"_2)))) = 5.021 * 10^(23)"molec."#

So, you know that your room contains **molecules** of chlorodifluoromethane and that volume of the room is said to be equal to

In order to determine how many molecules would be present in *cubic centimeters* to *cubic meters*

#250 color(red)(cancel(color(black)("cm"^3))) * "1 m"^3/(10^6color(red)(cancel(color(black)("cm"^3)))) = 2.5 * 10^(-4)"m"^3#

Finally, use the known concentration of molecules in the room to get the number of molecules present in the sample

#2.5 * 10^(-4) color(red)(cancel(color(black)("m"^3))) * (5.021 * 10^(23)"molec.")/(200color(red)(cancel(color(black)("m"^3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(6.3 * 10^(17)"molec.")color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**, but notice that you only have one sig fig for the volume of the room.