# Question #95250

##### 1 Answer

#### Answer:

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 **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 **for every**

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 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"#