The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, I ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$ \mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }} $$ In the case of very dilute aqueous solutions, a concentration of \(1.0 \mathrm{ppm}\) is equal to \(1.0 \mu \mathrm{g}\) of solute per \(1.0 \mathrm{mL},\) which equals 1.0 g solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. \(5.0 \mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

Step by step solution

01

Convert ppb to grams

Using the formula given in the problem, we can convert ppb to grams: $$ 5.0 \, \mathrm{ppb} = \frac{5.0 \, \mu \mathrm{g}}{1.0 \, \mathrm{L}} $$ To convert this to grams, multiply by \(10^{-9}\): $$ 5.0\, \mu \mathrm{g} \times 10^{-9} = 5.0 \times 10^{-9} \mathrm{g} $$

02

Calculate the number of moles of the solute

To find the number of moles, we need to divide the mass of the solute by its molar mass. The molar mass for \(Hg\) is \(200.6 \, \mathrm{g/mol}\): $$ \text{moles} = \frac{5.0 \times 10^{-9}\, \mathrm{g}}{200.6\, \mathrm{g/mol}} = 2.5 \times 10^{-11}\, \mathrm{mol} $$

03

Calculate the molarity of the solution

The solution has a volume of \(1.0 \, \mathrm{L}\), so we can calculate the molarity as: $$ \text{molarity} = \frac{2.5 \times 10^{-11}\, \mathrm{mol}}{1.0\, \mathrm{L}} = 2.5 \times 10^{-11}\, \mathrm{M} $$ b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\)

04

Convert ppb to grams

$$ 1.0\, \mathrm{ppb} = \frac{1.0\, \mu \mathrm{g}}{1.0\, \mathrm{L}} $$ $$ 1.0\, \mu \mathrm{g} \times 10^{-9} = 1.0 \times 10^{-9} \mathrm{g} $$

05

Calculate the number of moles of the solute

The molar mass for \(\mathrm{CHCl}_{3}\) is approximately \(119.4\, \mathrm{g/mol}\): $$ \text{moles} = \frac{1.0 \times 10^{-9}\, \mathrm{g}}{119.4\, \mathrm{g/mol}} = 8.4 \times 10^{-12}\, \mathrm{mol} $$

06

Calculate the molarity of the solution

$$ \text{molarity} = \frac{8.4 \times 10^{-12}\, \mathrm{mol}}{1.0\, \mathrm{L}} = 8.4 \times 10^{-12}\, \mathrm{M} $$ c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\)

07

Convert ppm to grams

$$ 10.0\, \mathrm{ppm} = \frac{10.0\, \mathrm{mg}}{1.0\, \mathrm{L}} $$ $$ 10.0\, \mathrm{mg} \times 10^{-3} = 1.0 \times 10^{-2} \mathrm{g} $$

08

Calculate the number of moles of the solute

The molar mass for \(As\) is approximately \(74.9\, \mathrm{g/mol}\): $$ \text{moles} = \frac{1.0 \times 10^{-2}\, \mathrm{g}}{74.9\, \mathrm{g/mol}} = 1.3 \times 10^{-4}\, \mathrm{mol} $$

09

Calculate the molarity of the solution

$$ \text{molarity} = \frac{1.3 \times 10^{-4}\, \mathrm{mol}}{1.0\, \mathrm{L}} = 1.3 \times 10^{-4}\, \mathrm{M} $$ d. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

10

Convert ppm to grams

$$ 0.10\, \mathrm{ppm} = \frac{0.10\, \mathrm{mg}}{1.0\, \mathrm{L}} $$ $$ 0.10\, \mathrm{mg} \times 10^{-3} = 1.0 \times 10^{-4} \mathrm{g} $$

11

Calculate the number of moles of the solute

The molar mass for \(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\) is approximately \(354.5\, \mathrm{g/mol}\): $$ \text{moles} = \frac{1.0 \times 10^{-4}\, \mathrm{g}}{354.5\, \mathrm{g/mol}} = 2.8 \times 10^{-7}\, \mathrm{mol} $$

12

Calculate the molarity of the solution

$$ \text{molarity} = \frac{2.8 \times 10^{-7}\, \mathrm{mol}}{1.0\, \mathrm{L}} = 2.8 \times 10^{-7}\, \mathrm{M} $$

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