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Sagot :
Answer:
3.39x10⁻⁹ M
Explanation:
Let's assume we have 1 mL of the stream. With a density of 1.0 g/mL, we would have 1 g of the stream.
Now we calculate the mass of mercury present in 1 g of the stream, using its given concentration (0.68 ppb):
- ppb = ng Mercury / g stream
- 0.68 ppb = ng Hg / 1 g
- Mass of mercury = 0.68 ng
Now we convert that mass into moles, using mercury's atomic mass:
- 0.68 ng = 0.68 x 10⁻⁹ g
- 0.68 x 10⁻⁹ g ÷ 200.59 g/mol = 3.39x10⁻¹² mol
Finally we calculate the molarity of mercury, as we know that there are 3.39x10⁻¹² Hg moles in 1 mL of the stream:
- 1 mL = 0.001 L
- Molarity = 3.39x10⁻¹² mol / 0.001 L = 3.39x10⁻⁹ M
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