Chapter 21, Problem 21.16QP

(a)

Interpretation Introduction

Interpretation: The concentrations of each of the given elements are to be expressed in parts per billion.

Concept introduction: In human body, concentrations of ultratrace elements are even lower than those of trace essential elements. Therefore, they are sometimes expressed in parts per billion.

To determine: The concentration of the given element in parts per billion.

Answer to Problem 21.16QP

Solution

The concentration is $\underset{\_}{6000ppb}$.

Explanation of Solution

Explanation

The given element is bromine, $\text{6}\text{mg}$ in $1\text{L}$.

The formula that relates the given quantities to ppm is,

$\text{1}\text{ppm}=1\text{mg/L}$

Hence, concentration in ppm is calculated by the formula,

$\text{Concentration}\text{in}\text{ppm}=\frac{\text{mg}\text{of}\text{solute}}{\text{L}\text{of}\text{solution}}$

The mass of solute is $\text{6}\text{mg}$ and volume of solution in $1\text{L}$.

Substitute the value of mass of solute and solution in the above formula.

$\begin{array}{c}\text{Concentration}\text{in}\text{ppm}=\frac{6\text{mg}}{1\text{L}}\\ =6\text{ppm}\end{array}$

The conversion of ppm to ppb is done as,

$1\text{ppm}=1000\text{ppb}$

Therefore, the conversion of $\text{6}\text{ppm}$ to ppb is done as,

$\begin{array}{c}\text{6}\text{ppm}=6\times 1000\text{ppb}\\ =\underset{\_}{6000ppb}\end{array}$

(b)

Interpretation Introduction

To determine: The concentration of the given element in parts per billion.

Answer to Problem 21.16QP

Solution

The concentration is $\underset{\_}{140ppb}$.

Explanation of Solution

Explanation

The given element is boron, $\text{0}\text{.014}\text{g/100}\text{kg}$.

The formula that relates the given quantities to ppm is,

$\text{1}\text{ppm}=1\frac{\text{mg}\text{of}\text{solute}}{\text{kg}\text{of}\text{solution}}$

Hence, concentration in ppm is calculated by the formula,

$\text{Concentration}\text{in}\text{ppm}=\frac{\text{mg}\text{of}\text{solute}}{\text{kg}\text{of}\text{solution}}$

The mass of solute is $\text{0}\text{.014}\text{g}$ and mass of solution in $1000\text{kg}$.

The conversion of $\text{g}$ to $\text{mg}$ is done as,

$1\text{g}=1000\text{mg}$

Therefore, the conversion of $\text{0}\text{.014}\text{g}$ to $\text{mg}$ is done as,

$\begin{array}{c}\text{0}\text{.014}\text{g}=\text{0}\text{.014}\times 1000\text{mg}\\ =14\text{mg}\end{array}$

Substitute the value of mass of solute and solution in the above formula.

$\begin{array}{c}\text{Concentration}\text{in}\text{ppm}=\frac{14\text{mg}}{100\text{kg}}\\ =0.14\text{ppm}\end{array}$

The conversion of ppm to ppb is done as,

$1\text{ppm}=1000\text{ppb}$

Therefore, the conversion of $0.14\text{ppm}$ to ppb is done as,

$\begin{array}{c}0.14\text{ppm}=0.14\times 1000\text{ppb}\\ =\underset{\_}{140ppb}\end{array}$

(c)

Interpretation Introduction

To determine: The concentration of the given element in parts per billion.

Answer to Problem 21.16QP

Solution

The concentration is $\underset{\_}{71ppb}$.

Explanation of Solution

Explanation

The given element is selenium, $\text{5}\text{.0}\text{mg/70}\text{kg}$.

The formula that relates the given quantities to ppm is,

$\text{1}\text{ppm}=1\frac{\text{mg}\text{of}\text{solute}}{\text{kg}\text{of}\text{solution}}$

Hence, concentration in ppm is calculated by the formula,

$\text{Concentration}\text{in}\text{ppm}=\frac{\text{mg}\text{of}\text{solute}}{\text{kg}\text{of}\text{solution}}$

The mass of solute is $\text{5}\text{.0}\text{mg}$ and mass of solution in $70\text{kg}$.

Substitute the value of mass of solute and solution in the above formula.

$\begin{array}{c}\text{Concentration}\text{in}\text{ppm}=\frac{5\text{mg}}{70\text{kg}}\\ =\text{0}\text{.071}\text{ppm}\end{array}$

The conversion of ppm to ppb is done as,

$1\text{ppm}=1000\text{ppb}$

Therefore, the conversion of $\text{0}\text{.071}\text{ppm}$ to ppb is done as,

$\begin{array}{c}\text{0}\text{.071}\text{ppm}=0.071\times 1000\text{ppb}\\ =\underset{\_}{71ppb}\end{array}$

Conclusion

The concentration of selenium, $\text{5}\text{.0}\text{mg/70}\text{kg}$ is $\underset{\_}{71ppb}$.