Question

Question #95250

Answer 1

Here's what I got.

Explanation:

We use parts per million, or ppm, to denote very, very low concentrations of solutes in a given solution or mixture.

More specifically, we say that a concentration of `"1 ppm"` corresponds to `1` part solute for every

`10^6 = 1,000,000`

parts of solution.

In your case, you know that the sample has a total mass of `"650 g"` and that carbon dioxide has a concentration of `"400 ppm"`. This tells you that for every `10^6` `"g"` of air, you get `"400 g"` of carbon dioxide.

Since the sample has the same composition throughout--air is a homogeneous solution--you can use its ppm concentration to figure out how many grams of carbon dioxide would be present in `"650 g"` of air.

`650 color(red)(cancel(color(black)("g air"))) * overbrace("400 g CO"_2/(10^6color(red)(cancel(color(black)("g air")))))^(color(blue)("= 400 ppm CO"_2)) = "0.26 g CO"_2`

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the ppm concentration of the sample.

Adjusted to the correct number of sig figs, you would have

`"mass CO"_2 = "30 g"`