Question

`"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 xx 20^23 "mol"^-1`

Answer 1

`6.3 * 10^(17)`

Explanation:

Your strategy here will be to use the molar mass of chlorodifluoromethane, `"CHClF"_2`, and Avogadro's number to calculate how many molecules of you have in that `7.21 * 10^(-2)"kg"` sample.

Chlorodifluoromethane has a molar mass of `"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 * 10^(23)` molecules of chlorodifluoromethane and that volume of the room is said to be equal to `"200 m"^3`.

In order to determine how many molecules would be present in `"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.