Explanation:
It is known that 1 ppb equals to 1 [tex]\mu g/L[/tex].
Therefore, 6.92 ppb is 6.92 [tex]\mu g[/tex] in 1 liter of sample.
Hence, we will calculate the moles of Anthracene as follows.
[tex]6.92 \mu g PHA \times \frac{1 mg PHA}{1000 \mu g PHA} \times \frac{1 g PHA}{1000 mg PHA} \times \frac{1 mol PHA}{178.23 g/mol}[/tex]
= [tex]0.038 \times 10^{-6}[/tex] mol PHA (Anthracene)
As we assume that volume of the solution is 1 L. So, calculate its molarity as follows.
Molarity = [tex]\frac{\text{no. of moles}}{Volume}[/tex]
= [tex]\frac{0.038 \times 10^{-6}}{1 L}[/tex]
= [tex]3.48 \times 10^{-8}[/tex] M
Thus, we can conclude that molar concentration of the polycyclic aromatic hydrocarbon (PAH) is [tex]3.48 \times 10^{-8}[/tex] M.
The molarity of anthracene is 3.88 × 10^-8 M.
It is common to express the concentration of contaminants in water in units of parts per million or parts per billion. In the context of this problem, the concentration of anthracene is 6.92 ppb. Note that ppb is the same unit as μg/L.
We can rewrite the concentration as 6.92 μg/L. Hence, 6.92 μg of anthracene is contained in 1L of solution.
Numbr of moles of anthracene = 6.92 × 10^-6 g/178.23 g/mol = 3.88 × 10^-8 moles
Since the total volume present is 1L of solution;
Molarity = number of moles / volume
Molarity = 3.88 × 10^-8 moles/1 L
Molarity = 3.88 × 10^-8 M
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