A #"5 gallon"# volume of benzene is accidentally added to an aquifer whose volume is #3.79xx10^8*L#. If homogeneous mixing is assumed, what is the #"ppm concentration"# with respect to benzene?

Question

A #"5 gallon"# volume of benzene is accidentally added to an aquifer whose volume is #3.79xx10^8*L#.  If homogeneous mixing is assumed, what is the #"ppm concentration"# with respect to benzene?

Naomi Aronson · Accepted Answer

Whoops! I wonder if this an actual scenario?

Since both the metric and US (?) systems of volume measurement have been applied, this question is certain to be problematic. We will assume that the benzene is uniformly mixed throughout the aquifer. Now, 1 US gallon #-=# #3.78541# #L#. So volume of aquifer #=# #10^8* cancel(gallons)# #xx# #3.78541*L*cancel(gallon^-1)# #=# #3.79 xx 10^8 *L#. And #5 *cancel(gallons)xx3.78541# #L*cancel(gallons^-1)# #=# #18.9*L# is the volume of benzene that some klutz introduced. Benzene has a density of #0.874# #g*mL^-1#. So mass of benzene #=# #18.9*cancelLxx10^3cancel(mL)*cancelL^-1xx0.874*g*cancel(mL^-1)# #=# #16,518.6 *g# benzene. So we've got (finally) #(16,518.6 *g)/(3.79 xx 10^8 L)# #=# #4.4xx10^-5g*L^-1# #=# #4.4xx10^-2*mg*L^-1#. And this assumes homogeneous mixing, which might not be the case. This is well under #"1 ppm"#. But please check my figures. I will not guarantee the accuracy. And who said, #"the solution to pollution is dilution"?#

Olivia Anton · Answer

undefined To find the concentration of benzene in parts per million (ppm), first convert the volume of benzene from gallons to liters. Then, divide the amount of benzene in liters by the total volume of the aquifer in liters and multiply by 10^6.

ppm concentration of benzene = (volume of benzene in liters / total volume of aquifer in liters) * 10^6