Question 7ba62

Nov 6, 2015

$\text{355 mg}$

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.

$\textcolor{b l u e}{\text{ppm" = "mass of solute"/"mass of solvent} \times {10}^{6}}$

You can assume that one liter of rain water has a density of approximately ${\text{1 g/cm}}^{3}$. Keeping in mind that you have

${\text{1 L" = "1 dm"^3 = 10^3"cm}}^{3}$

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.

$\text{ppm" = m_(CO_2)/m_"water} \times {10}^{6}$

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

${m}_{C {O}_{2}} = \frac{\text{ppm" xx m_"water}}{10} ^ 6$

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

If you want, you can express this value in miligrams

${m}_{C {O}_{2}} = \textcolor{g r e e n}{\text{355 mg}}$