Question

Question #7ba62

Answer 1

`"355 mg"`

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 `10^6` parts solvent.

`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 g/cm"^3`. Keeping in mind that you have

`"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 `10^6`, you get the concentration in ppm.

`"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")`