# 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?

Nov 18, 2015

#### Answer:

Whoops! I wonder if this an actual scenario?

#### Explanation:

This question is bound to be problematic because metric and US (?) systems of volume measurement have been used. We assume that the benzene is homogeneously mixed through the aquifer.

Now, 1 US gallon $\equiv$ $3.78541$ $L$.

So volume of aquifer $=$ ${10}^{8} \cdot \cancel{g a l l o n s}$ $\times$ $3.78541 \cdot L \cdot \cancel{g a l l o {n}^{-} 1}$ $=$ $3.79 \times {10}^{8} \cdot L$.

And $5 \cdot \cancel{g a l l o n s} \times 3.78541$ $L \cdot \cancel{g a l l o n {s}^{-} 1}$ $=$ $18.9 \cdot L$ is the volume of benzene that some klutz introduced. Benzene has a density of $0.874$ $g \cdot m {L}^{-} 1$.

So mass of benzene $=$ $18.9 \cdot \cancel{L} \times {10}^{3} \cancel{m L} \cdot {\cancel{L}}^{-} 1 \times 0.874 \cdot g \cdot \cancel{m {L}^{-} 1}$ $=$ $16 , 518.6 \cdot g$ benzene.

So we've got (finally) $\frac{16 , 518.6 \cdot g}{3.79 \times {10}^{8} L}$ $=$ $4.4 \times {10}^{-} 5 g \cdot {L}^{-} 1$ $=$ $4.4 \times {10}^{-} 2 \cdot m g \cdot {L}^{-} 1$. And this assumes homogeneous mixing, which might not be the case.

This is well under $\text{1 ppm}$. But please check my figures. I will not guarantee the accuracy. And who said, "the solution to pollution is dilution"?