Question

10.30 mg of $O_2$ is dissolved into a liter of sea water of density 1.03 g/mL. The concentration of $O_2$ in ppm is

• A. 8
• B. 12
• C. 15
• D. 10

Answer 1

Answer: (D)

$ppm=\frac{W_{\text{Solute }}}{W_{\text{Solution }}}\times 100$
Using the density of the solution and its volume $\left(1L=1000mL=1000c{m}^3\right)$, the weight of the solution can be calculated.
$W_{\text{solution }}=1.03\times 1000=1030g$
Thus, ppm $\frac{10.3\times 10^{-3}g}{1030g}\times 10^6=10ppm$