# "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?

## The Avogadro constant is $6.02 \times {20}^{23} {\text{mol}}^{-} 1$

Aug 4, 2016

#### Answer:

$6.3 \cdot {10}^{17}$

#### Explanation:

Your strategy here will be to use the molar mass of chlorodifluoromethane, ${\text{CHClF}}_{2}$, and Avogadro's number to calculate how many molecules of you have in that $7.21 \cdot {10}^{- 2} \text{kg}$ sample.

Chlorodifluoromethane has a molar mass of ${\text{86.47 g mol}}^{- 1}$, so start by converting your sample from 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 $5.021 \cdot {10}^{23}$ molecules of chlorodifluoromethane and that volume of the room is said to be equal to ${\text{200 m}}^{3}$.

In order to determine how many molecules would be present in ${\text{250 cm}}^{3}$ of air, convert the volume of the sample from 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.