# A 300*lb mass of a chemical is accidentally spilled into a dam whose volume is 1.362xx10^4*m^3... What is the "ppm" concentration?

Jun 25, 2017

Whoops.....well first we have to have kosher units.......We ASSUME the chemical is homogeneously mixed in the dam, and we finally get a volume of over 10 million litres.........

#### Explanation:

$\text{1 ppm} = 1 \cdot m g \cdot {L}^{-} 1 = {10}^{-} 3 \cdot g \cdot {L}^{-} 1$. At $\text{ppm}$ levels of concentration we don't really have to worry about solution density.

We gots $300 \cdot l b$ of stuff, i.e.

$300 \cdot l b \times 454 \cdot g \cdot l {b}^{-} 1 \times {10}^{3} \cdot m g \cdot {g}^{-} 1 = 1.362 \times {10}^{8} \cdot m g$

And now we can write out the quotient..........

$\text{mass"/"volume"="concentration} = 10 \cdot m g \cdot {L}^{-} 1$.........

So $\text{volume"="mass"/"concentration} = \frac{1.362 \times {10}^{8} \cdot m g}{10 \cdot m g \cdot {L}^{-} 1}$ and we gets an answer in $L$, which we then convert to ${m}^{3}$

$= 1.362 \times {10}^{7} \cdot L = 1.362 \times {10}^{4} \cdot {m}^{3}$, not an insignificant volume......

All I have done is to use the unit conversions, and know that $\text{milli} = {10}^{-} 3$.......I agree that it is not wholly straightforward, but practice is the key.