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      Roadbed Guy, mmacdDE, Jaimas

      It looks like you're trying to do a calculation where you've confused mass fractions with volumetric fractions.

      The EPA maximum contaminant level for benzene in drinking water is 0.005 mg/L, which isn't a volumetric fraction.   Moreover, the bulk density of water is different than
      benzene.

      EPA drinking water standard for benzene at:

      http://water.epa.gov/...

      The postulation is 54 gallons of benzene was contained in the spill and all of it was available for contamination of water (which it isn't because of rapid evaporation of light molecular weight compounds from the crude mixture).

      The bulk density of benzene is 0.879 grams/ml
      see    http://www.epa.gov/...

      54 gallons C6H6  = (3.785 L/gal)54 gal  =  204 Liters

      204 Liters = 204000 ml C6H6

      (0.879 grams C6H6/ml) * 204000 ml C6H6 =

      ==   179316 grams of C6H6 would be contained in 54 gallons of 100% benzene.

      How much water would this amount of benzene contaminate to the level of the EPA MCL for benzene if it were 100% available available for water contamination?

      X = volume of water contaminated by 179316 grams of benzene when the aqueous concentration is at the MCL of 0.005 mg/Liter

      (0.005 mg C6H6/Liter H20) X = (179316) grams * 1E3 mg/gram

      X =  (179316) grams C6H6 * 1E3 mg/gram
              ---------------------------------------
                 0.005 mg C6H6/Liter H20

      X =   3.6E10 Liters of H20

      X =    9.5E09 Gallons of H20  (3X higher than your result)

      X =    1.1E09 cubic feet of H20

      This is a CUBE of water 1032 feet high.

      Assuming that all the spilled benzene would be 100%  available to contaminate water, but it will not be because of evaporation.

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