Chapter 4, Problem 31E

(a)

Interpretation Introduction

Interpretation: The molarity of the $5.0\text{ppb}$ of $\text{Hg}$ in ${\text{H}}_2\text{O}$ is to be calculated.

Concept introduction:The molarity of any substance is obtained from the moles of substance to the volume of the solution.

The formula to calculate the molarity is as follows.

  $\text{Molarity}=\frac{\text{Moles}\left(\text{mol}\right)}{\text{Volume}\left(\text{L}\right)}$

(b)

Interpretation Introduction

Interpretation:The molarity of the $1.0\text{ppb}$ of ${\text{CHCl}}_3$ in ${\text{H}}_2\text{O}$ is to be calculated.

Concept introduction:The molarity of any substance is obtained from the moles of substance to the volume of the solution.

The formula to calculate the molarity is as follows.

  $\text{Molarity}=\frac{\text{Moles}\left(\text{mol}\right)}{\text{Volume}\left(\text{L}\right)}$

(c)

Interpretation Introduction

Interpretation:The molarity of the $10.0\text{ppm}$ of $\text{As}$ in ${\text{H}}_2\text{O}$ is to be calculated.

Concept introduction:The molarity of any substance is obtained from the moles of substance to the volume of the solution.

The formula to calculate the molarity is as follows.

  $\text{Molarity}=\frac{\text{Moles}\left(\text{mol}\right)}{\text{Volume}\left(\text{L}\right)}$

(d)

Interpretation Introduction

Interpretation:The molarity of the $0.10\text{ppm}$ of ${\text{C}}_{14}{\text{H}}_9{\text{Cl}}_5$ that is DDT in ${\text{H}}_2\text{O}$ is to be calculated.

Concept introduction:The molarity of any substance is obtained from the moles of substance to the volume of the solution.

The formula to calculate the molarity is as follows.

  $\text{Molarity}=\frac{\text{Moles}\left(\text{mol}\right)}{\text{Volume}\left(\text{L}\right)}$