# Question #65e0c

Sep 14, 2017

Well $\text{1 ppm} \equiv 1 \cdot m g \cdot {L}^{-} 1$, and so our $\text{ppm}$ concentration will be rather large.....

#### Explanation:

And we gots.......$\frac{36.2 \cdot g}{\left(100 - 36.2\right) \times {10}^{-} 3 \cdot L} = 567.4 \cdot g \cdot {L}^{-} 1$

$\equiv 567.4 \cdot g \times {10}^{3} \cdot m g \cdot {g}^{-} 1 \cdot {L}^{-} 1 \equiv 567398 \cdot m g \cdot {L}^{-} 1$

$\equiv 567398 \cdot \text{ppm}$....

And here we have ASSUMED (reasonably), that the volume of the solution would be approx. $64 \cdot m L$.

I would also check your quoted concentration. This site quotes a solubility of $359 \cdot g \cdot {L}^{-} 1$.....