Chapter 13, Problem 56E

Interpretation Introduction

(a)

Interpretation:

The two pairs of unit factors for $0.150M\text{KBr}$ are to be stated.

Concept introduction:

Molarity of a solution is the amount of solute dissolved in a $\text{1000}\text{mL}$ solution. It is also defined as the number of moles of solute per volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 56E

The two pairs of unit factors for $0.150M\text{KBr}$ are shown below.

$\frac{\text{0}\text{.150}\text{mol}\text{KBr}}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.150}\text{mol}\text{KBr}}$ and $\frac{\text{0}\text{.150}\text{mol}\text{KBr}}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.150}\text{mol}\text{KBr}}$

Explanation of Solution

Unit factors are numeric expressions that are used to convert one unit into another unit of a parameter. For example, the unit of molarity is $M$ and to convert into $\text{mol}/\text{L}$, unit factors are required that is $\text{1}\text{mol}$ of solute is present in $\text{1L}$ of solution.

Therefore, the pair of unit factor is stated below.

$\frac{\text{0}\text{.150}\text{mol}\text{KBr}}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.150}\text{mol}\text{KBr}}$

The relation between $\text{L}$ and $\text{mL}$ is given below.

$\text{1}\text{L}=\text{1000}\text{mL}$

Therefore, the pair of another unit factor is shown below.

$\frac{\text{0}\text{.150}\text{mol}\text{KBr}}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.150}\text{mol}\text{KBr}}$

Conclusion

The two pairs of unit factors for $0.150M\text{KBr}$ have been rightfully stated.

Interpretation Introduction

(b)

Interpretation:

The two pairs of unit factors for $0.150M\text{Ca}{\left({\text{NO}}_3\right)}_2$ are to be stated.

Concept introduction:

Molarity of a solution is the amount of solute dissolved in a $\text{1000}\text{mL}$ solution. It is also defined as the number of moles of solute per volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 56E

The two pairs of unit factors for $0.150M\text{Ca}{\left({\text{NO}}_3\right)}_2$ are shown below.

$\frac{\text{0}\text{.150}\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.150}\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}$$\frac{\text{0}\text{.150}\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.150}\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}$

Explanation of Solution

Unit factors are numeric expressions that are used to convert one unit into another unit of a parameter. For example, the unit of molarity is $M$ and to convert into $\text{mol}/\text{L}$, unit factors are required that is $\text{1}\text{mol}$ of solute is present in $\text{1L}$ of solution.

Therefore, the pair of unit factor is stated below.

$\frac{\text{0}\text{.150}\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.150}\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}$

The relation between $\text{L}$ and $\text{mL}$ is given below.

$\text{1}\text{L}=\text{1000}\text{mL}$

Therefore, the pair of another unit factor is shown below.

$\frac{\text{0}\text{.150}\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.150}\text{mol}\text{Ca}{\left({\text{NO}}_3\right)}_2}$

Conclusion

The two pairs of unit factors for $0.150M\text{Ca}{\left({\text{NO}}_3\right)}_2$ have been rightfully stated.

Interpretation Introduction

(c)

Interpretation:

The two pairs of unit factors for $0.333M\text{Sr}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ are to be stated.

Concept introduction:

Molarity of a solution is the amount of solute dissolved in a $\text{1000}\text{mL}$ solution. It is also defined as the number of moles of solute per volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 56E

The two pairs of unit factors for $0.333M\text{Sr}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ are shown below.

$\frac{\text{0}\text{.333}\text{mol}\text{Sr}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.333}\text{mol}\text{Sr}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}}$$\frac{\text{0}\text{.333}\text{mol}\text{Sr}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.333}\text{mol}\text{Sr}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}}$

Explanation of Solution

Unit factors are numeric expressions that are used to convert one unit into another unit of a parameter. For example, the unit of molarity is $M$ and to convert into $\text{mol}/\text{L}$, unit factors are required that is $\text{1}\text{mol}$ of solute is present in $\text{1L}$ of solution.

Therefore, the pair of unit factor is stated below.

$\frac{\text{0}\text{.333}\text{mol}\text{Sr}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.333}\text{mol}\text{Sr}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}}$

The relation between $\text{L}$ and $\text{mL}$ is given below.

$\text{1}\text{L}=\text{1000}\text{mL}$

Therefore, the pair of another unit factor is shown below.

$\frac{\text{0}\text{.333}\text{mol}\text{Sr}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}}{\text{1000}\text{mL}\text{solution}},\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.333}\text{mol}\text{Sr}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}}$

Conclusion

The two pairs of unit factors for $0.333M\text{Sr}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ have been rightfully stated.

Interpretation Introduction

(d)

Interpretation:

The two pairs of unit factors for $0.333M{\text{NH}}_{\text{4}}\text{Cl}$ are to be stated.

Concept introduction:

Molarity of a solution is the amount of solute dissolved in a $\text{1000}\text{mL}$ solution. It is also defined as the number of moles of solute per volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 56E

The two pairs of unit factors for $0.333M{\text{NH}}_{\text{4}}\text{Cl}$ are shown below.

$\frac{\text{0}\text{.333}\text{mol}{\text{NH}}_{\text{4}}\text{Cl}}{\text{1}\text{L}\text{solution}},\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.333}\text{mol}{\text{NH}}_{\text{4}}\text{Cl}}$ and $\frac{\text{0}\text{.333}\text{mol}{\text{NH}}_{\text{4}}\text{Cl}}{\text{1000}\text{mL}\text{solution}},\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.333}\text{mol}{\text{NH}}_{\text{4}}\text{Cl}}$

Explanation of Solution

Unit factors are numeric expressions that are used to convert one unit into another unit of a parameter. For example, the unit of molarity is $M$ and to convert into $\text{mol}/\text{L}$, unit factors are required that is $\text{1}\text{mol}$ of solute is present in $\text{1L}$ of solution.

Therefore, the pair of unit factor is stated below.

$\frac{\text{0}\text{.333}\text{mol}{\text{NH}}_{\text{4}}\text{Cl}}{\text{1}\text{L}\text{solution}},\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.333}\text{mol}{\text{NH}}_{\text{4}}\text{Cl}}$

The relation between $\text{L}$ and $\text{mL}$ is given below.

$\text{1}\text{L}=\text{1000}\text{mL}$

Therefore, the pair of another unit factor is shown below.

$\frac{\text{0}\text{.333}\text{mol}{\text{NH}}_{\text{4}}\text{Cl}}{\text{1000}\text{mL}\text{solution}}$ and $\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.333}\text{mol}{\text{NH}}_{\text{4}}\text{Cl}}$

Conclusion

The two pairs of unit factors for $0.333M{\text{NH}}_{\text{4}}\text{Cl}$ have been rightfully stated.