Chapter 4, Problem 121CP

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every 10⁶ parts of solution. Mathematically, by mass:

$\text{ppm=}\frac{\text{μg}\text{solute}}{\text{g}\text{solution}}\text{=}\frac{\text{mg}\text{solute}}{\text{kg}\text{solution}}$

In the case of very dilute aqueous solutions, a concentration of 1.0 ppm is equal to 1.0 µg of solute per 1.0 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 ppb Hg in H₂O

b. 1.0 ppb CHCl₃ in H₂O

c. 10.0 ppm As in H₂O

d. 0.10 ppm DDT (C₁₄H₉Cl₅) in H₂O

Interpretation Introduction

Interpretation: The molarity of the aqueous solutions has to be calculated.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

Answer to Problem 121CP

$\text{Molarity}\text{of}\text{5}\text{.0ppb}\text{Hg}\text{in}\text{water=2}{\text{.5×10}}^{\text{-8}}\text{M}\text{Hg}$

Explanation of Solution

Record the given data

Mass of solute= $\text{5}\text{.0ppb Hg}$

Volume of solution= $\text{1}\text{.0mL}$

The mass of the solute and volume of solution are recorded as shown above.

To calculate the molarity of $\text{5}\text{.0ppb}\text{Hg}\text{in}\text{water}$

Molar mass of Hg= $\text{200}\text{.6}\text{g}$

$\begin{array}{l}\text{5}\text{.0}\text{ppb}\text{Hg}\text{in}\text{water=}\frac{\text{5}\text{.0}\text{ng}\text{Hg}}{\text{g}\text{soln}}\text{=}\frac{\text{5}{\text{.0×10}}^{\text{-9}}\text{g}\text{Hg}}{\text{mL}\text{soln}}\\ \frac{\text{5}{\text{.0×10}}^{\text{-9}}\text{g}\text{Hg}}{\text{mL}}\text{×}\frac{\text{1}\text{mol}\text{Hg}}{\text{200}\text{.6}\text{g}\text{Hg}}\text{×}\frac{\text{1000}\text{mL}}{\text{L}}\text{=2}{\text{.5×10}}^{\text{-8}}\text{M}\text{Hg}\end{array}$

$\text{Molarity}\text{of}\text{5}\text{.0ppb}\text{Hg}\text{in}\text{water=2}{\text{.5×10}}^{\text{-8}}\text{M}\text{Hg}$

The molarity of $\text{5}\text{.0ppb}$ of Hg in water is calculated by plugging in the values of moles with molar mass along with volume of the solution. The molarity of $\text{5}\text{.0ppb}$ of Hg in water is $\text{2}{\text{.5×10}}^{\text{-8}}\text{M}\text{Hg}$.

Interpretation Introduction

Interpretation: The molarity of the aqueous solutions has to be calculated.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

Answer to Problem 121CP

Molarity of $\text{1}\text{.0}\text{ppb}{\text{CHCl}}_{\text{3}}\text{in}{\text{H}}_{\text{2}}\text{O=8}{\text{.4×10}}^{\text{-9}}\text{M}{\text{CHCl}}_{\text{3}}$

Explanation of Solution

Record the given data

Mass of solute= $\text{1}\text{.0}\text{ppb}{\text{CHCl}}_{\text{3}}$

Volume of solution= $\text{1}\text{.0mL}$

The mass of the solute and volume of solution are recorded as shown above.

To calculate the molarity of $\text{1}\text{.0}\text{ppb}{\text{CHCl}}_{\text{3}}\text{in}{\text{H}}_{\text{2}}\text{O}$

Molar mass of ${\text{CHCl}}_{\text{3}}$ = $\text{119}\text{.73}\text{g}$

$\frac{\text{1}{\text{.0×10}}^{\text{-9}}\text{g}{\text{CHCl}}_{\text{3}}}{\text{mL}}\text{×}\frac{\text{1}\text{mol}{\text{CHCl}}_{\text{3}}}{\text{119}\text{.37}\text{g}{\text{CHCl}}_{\text{3}}}\text{×}\frac{\text{1000}\text{mL}}{\text{L}}\text{=8}{\text{.4×10}}^{\text{-9}}\text{M}{\text{CHCl}}_{\text{3}}$

Molarity of $\text{1}\text{.0}\text{ppb}{\text{CHCl}}_{\text{3}}\text{in}{\text{H}}_{\text{2}}\text{O=8}{\text{.4×10}}^{\text{-9}}\text{M}{\text{CHCl}}_{\text{3}}$

The molarity of $\text{1}\text{.0}\text{ppb}{\text{CHCl}}_{\text{3}}\text{in}{\text{H}}_{\text{2}}\text{O}$ is calculated by plugging in the values of moles with molar mass along with volume of the solution. The molarity of $\text{1}\text{.0}\text{ppb}{\text{CHCl}}_{\text{3}}\text{in}{\text{H}}_{\text{2}}\text{O}$ is $\text{8}{\text{.4×10}}^{\text{-9}}\text{M}{\text{CHCl}}_{\text{3}}$.

Interpretation Introduction

Interpretation: The molarity of the aqueous solutions has to be calculated.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

Answer to Problem 121CP

Molarity of $\text{10}\text{.0}\text{ppm}\text{As}\text{in}{\text{H}}_{\text{2}}\text{O=1}{\text{.3×10}}^{\text{-4}}\text{M}\text{As}$

Explanation of Solution

Record the given data

Mass of solute= $\text{10}\text{.00ppm}\text{As}$

Volume of solution= $\text{1}\text{.0mL}$

The mass of the solute and volume of solution are recorded as shown above.

To calculate the molarity of $\text{10}\text{.0}\text{ppm}\text{As}\text{in}{\text{H}}_{\text{2}}\text{O}$

Molar mass of As= $\text{74}\text{.92}\text{g}$

$\begin{array}{l}\text{10}\text{.0}\text{ppm}\text{As=}\frac{\text{10}\text{.0μg}\text{As}}{\text{g}\text{soln}}\text{=}\frac{\text{10}{\text{.0×10}}^{\text{-6}}\text{g}\text{As}}{\text{mL}\text{soln}}\\ \frac{\text{10}{\text{.0×10}}^{\text{-6}}\text{}\text{g}\text{As}}{\text{mL}}\text{×}\frac{\text{1}\text{mol}\text{As}}{\text{74}\text{.92}\text{g}\text{As}}\text{×}\frac{\text{1000}\text{mL}}{\text{L}}\text{=1}{\text{.33×10}}^{\text{-4}}\text{M}\text{As}\end{array}$

Molarity of $\text{10}\text{.0}\text{ppm}\text{As}\text{in}{\text{H}}_{\text{2}}\text{O}$= $\text{1}{\text{.33×10}}^{\text{-4}}\text{M}\text{As}$

The molarity of $\text{10}\text{.0}\text{ppm}\text{As}\text{in}{\text{H}}_{\text{2}}\text{O}$ is calculated by plugging in the values of moles with molar mass along with volume of the solution. The molarity of $\text{10}\text{.0}\text{ppm}\text{As}\text{in}{\text{H}}_{\text{2}}\text{O}$ is $\text{1}{\text{.33×10}}^{\text{-4}}\text{M}\text{As}$

Interpretation Introduction

Interpretation: The molarity of the aqueous solutions has to be calculated.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

Answer to Problem 121CP

Molarity of $\text{0}\text{.10}\text{ppm}{\text{DDT(C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{15}}\text{)}\text{in}{\text{H}}_{\text{2}}\text{O=2}{\text{.8×10}}^{\text{-7}}\text{M}\text{DDT}$

Explanation of Solution

Record the given data

Mass of solute= $\text{0}\text{.10}\text{ppm}{\text{DDT(C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{5}}\text{)}$

Volume of solution= $\text{1}\text{.0mL}$

The mass of the solute and volume of solution are recorded as shown above.

To calculate the molarity of $\text{0}\text{.10}\text{ppm}{\text{DDT(C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{5}}\text{)}$ in ${\text{H}}_{\text{2}}\text{O}$

Molar mass of ${\text{(C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{5}}\text{)}$= $\text{354}\text{.46}\text{g}$

$\frac{\text{0}{\text{.10×10}}^{\text{-6}}\text{g}\text{DDT}}{\text{mL}}\text{×}\frac{\text{1}\text{mol}\text{DDT}}{\text{354}\text{.46}\text{g}\text{DDT}}\text{×}\frac{\text{1000}\text{mL}}{\text{L}}\text{=2}{\text{.8×10}}^{\text{-7}}\text{M}\text{DDT}$

Molarity of $\text{0}\text{.10}\text{ppm}{\text{DDT(C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{5}}\text{)}$ in ${\text{H}}_{\text{2}}\text{O}$= $\text{2}{\text{.8×10}}^{\text{-7}}\text{M}\text{DDT}$

The molarity of $\text{0}\text{.10}\text{ppm}{\text{DDT(C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{5}}\text{)}$ in ${\text{H}}_{\text{2}}\text{O}$ is calculated by plugging in the values of moles with molar mass along with volume of the solution. The molarity of in ${\text{H}}_{\text{2}}\text{O}$ is $\text{2}{\text{.8×10}}^{\text{-7}}\text{M}\text{DDT}$.