# Question #7ba62

##### 1 Answer

#### Answer:

#### Explanation:

*Parts per million*, or ppm, are used when dealing with *very small* concentrations of solute.

In simple terms, a concentration of **one ppm** is equivalent to one part solute *per*

#color(blue)("ppm" = "mass of solute"/"mass of solvent" xx 10^6)#

You can assume that one liter of rain water has a density of approximately

#"1 L" = "1 dm"^3 = 10^3"cm"^3#

you can say that **one liter** of rain water will have a mass of

#1color(red)(cancel(color(black)("L"))) * "1 g"/(1color(red)(cancel(color(black)("cm"^3)))) * (10^3color(red)(cancel(color(black)("cm"^3))))/(1color(red)(cancel(color(black)("dm"^3)))) = 10^3"g"#

So, you know that if you take the ratio between the **mass of the solute** and the mass of the solvent, and **multiply the result** by

#"ppm" = m_(CO_2)/m_"water" xx 10^6#

This means that the mass of the solute can be determined by rearranging the equation

#m_(CO_2) = ("ppm" xx m_"water")/10^6#

#m_(CO_2) = (355 * 10^3"g")/10^6 = 355 * 10^(-3)"g"#

If you want, you can express this value in *miligrams*

#m_(CO_2) = color(green)("355 mg")#