Question

Calculate the molar concentration of the polycyclic aromatic hydrocarbon (pah), anthracene (178.23 g/mol), that was found in a well water sample at a concentration of 6.92 ppb. Assume the density of the water is 1.00 g/ml.

Answer 1

Explanation:

It is known that 1 ppb equals to 1 $\mu g/L$.

Therefore, 6.92 ppb is 6.92 $\mu g$ in 1 liter of sample.

Hence, we will calculate the moles of Anthracene as follows.

           $6.92 \mu g PHA \times \frac{1 mg PHA}{1000 \mu g PHA} \times \frac{1 g PHA}{1000 mg PHA} \times \frac{1 mol PHA}{178.23 g/mol}$

          = $0.038 \times 10^{-6}$ mol PHA (Anthracene)

As we assume that volume of the solution is 1 L. So, calculate its molarity as follows.

              Molarity = $\frac{\text{no. of moles}}{Volume}$

                            = $\frac{0.038 \times 10^{-6}}{1 L}$

                           = $3.48 \times 10^{-8}$ M

Thus, we can conclude that molar concentration of the polycyclic aromatic hydrocarbon (PAH) is $3.48 \times 10^{-8}$ M.

Answer 2

solution:

Calculate the molar concentration of the polycyclic aromatic hydrocarbon(PHA)(178.23g/mol),that was found in a well water sample at a concentration of 6.21ppb.Assume the density of the water is 1.00mg/mL$ppb corresponds to 1\mu g in 1L of sample \\the 6.21 ppb is 6.21\mu g in 1L of sample\\6.21\mu g PHA\times\frac{1mg PHA}{1000\mu g PHA}\times\frac{1g PHA}{1000\mu mg PHA}\times\frac{1mol. PHA}{178.23g PHA}\\3.48\times10^-8mol PHA\\and if we have 1L of solution\\m=\frac{moles}{v(L)}=\frac{3.48\times10^-8mol}{1L}\\3.348\times^10-8 M$