Solution Exercise 1

One milligram per liter of DOC divided by the atomic weight of carbon (12) times 1,000.

\displaystyle \frac{1\frac{\textrm{mg}}{\textrm{L}}}{12\frac{\textrm{g}}{\textrm{mol}}\textrm{1,000}\frac{\textrm{mg}}{\textrm{g}}}\;\textrm{1,000,000}\frac{\mu \textrm{mol}}{\textrm{mol}}=83.3\frac{\mu \textrm{mol}}{\textrm{L}}

Return to Exercise 1

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Dissolved Organic Carbon in Groundwater Systems Copyright © 2022 by Francis H. Chapelle. All Rights Reserved.