10.30 mg of \(\mathrm{O_2}\) is dissolved into a liter of sea water of density 1.03 g/mL. The concentration of \(\mathrm{O_2}\) in ppm is__________.
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Correct Answer: 10
Solution and Explanation
Mass of \(\mathrm{O_2}\) = 10.30 mg = \(10.30\times10^{-3}\) g.
ppm = \(\frac{\text{mass of solute}}{\text{mass of solution}}\times10^6 = \frac{10.30\times10^{-3}}{1030}\times10^6\).
\(= \frac{10300}{1030} = 10\).
Answer: 10
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