Chapter 13, Problem 55E

Interpretation Introduction

(a)

Interpretation:

The two pairs of unit factors for $0.100M\text{LiI}$ 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 55E

The two pairs of unit factors for $0.100M\text{LiI}$ are shown below.

$\frac{\text{0}\text{.100}\text{mol}\text{LiI}}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.100}\text{mol}\text{LiI}}$ and $\frac{\text{0}\text{.100}\text{mol}\text{LiI}}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.100}\text{mol}\text{LiI}}$

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.

$\frac{\text{0}\text{.100}\text{mol}\text{LiI}}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.100}\text{mol}\text{LiI}}$

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.

$\frac{\text{0}\text{.100}\text{mol}\text{LiI}}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.100}\text{mol}\text{LiI}}$

Conclusion

The two pairs of unit factors for $0.100M\text{LiI}$ have been rightfully stated.

Interpretation Introduction

(b)

Interpretation:

The two pairs of unit factors for $0.100M{\text{NaNO}}_3$ 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 55E

The two pairs of unit factors for $0.100M{\text{NaNO}}_3$ are shown below.

$\frac{\text{0}\text{.100}\text{mol}{\text{NaNO}}_3}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.100}\text{mol}{\text{NaNO}}_3}$ and $\frac{\text{0}\text{.100}\text{mol}{\text{NaNO}}_3}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.100}\text{mol}{\text{NaNO}}_3}$

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.

$\frac{\text{0}\text{.100}\text{mol}{\text{NaNO}}_3}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.100}\text{mol}{\text{NaNO}}_3}$

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.

$\frac{\text{0}\text{.100}\text{mol}{\text{NaNO}}_3}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.100}\text{mol}{\text{NaNO}}_3}$

Conclusion

The two pairs of unit factors for $0.100M{\text{NaNO}}_3$ have been rightfully stated.

Interpretation Introduction

(c)

Interpretation:

The two pairs of unit factors for $0.500M{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ 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 55E

The two pairs of unit factors for $0.500M{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ are shown below.

$\frac{\text{0}\text{.500}\text{mol}{\text{K}}_2{\text{CrO}}_4}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.500}\text{mol}{\text{K}}_2{\text{CrO}}_4}$ and $\frac{\text{0}\text{.500}\text{mol}{\text{K}}_2{\text{CrO}}_4}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.500}\text{mol}{\text{K}}_2{\text{CrO}}_4}$

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.

$\frac{\text{0}\text{.500}\text{mol}{\text{K}}_2{\text{CrO}}_4}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.500}\text{mol}{\text{K}}_2{\text{CrO}}_4}$

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.

$\frac{\text{0}\text{.500}\text{mol}{\text{K}}_2{\text{CrO}}_4}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.500}\text{mol}{\text{K}}_2{\text{CrO}}_4}$

Conclusion

The two pairs of unit factors for $0.500M{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ have been rightfully stated.

Interpretation Introduction

(d)

Interpretation:

The two pairs of unit factors for $0.500M{\text{ZnSO}}_{\text{4}}$ 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 55E

The two pairs of unit factors for $0.500M{\text{ZnSO}}_{\text{4}}$ are shown below.

$\frac{\text{0}\text{.500}\text{mol}{\text{ZnSO}}_4}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.500}\text{mol}{\text{ZnSO}}_4}$ and $\frac{\text{0}\text{.500}\text{mol}{\text{ZnSO}}_4}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.500}\text{mol}{\text{ZnSO}}_4}$

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.

$\frac{\text{0}\text{.500}\text{mol}{\text{ZnSO}}_4}{\text{1}\text{L}\text{solution}}\text{,}\frac{\text{1}\text{L}\text{solution}}{\text{0}\text{.500}\text{mol}{\text{ZnSO}}_4}$

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.

$\frac{\text{0}\text{.500}\text{mol}{\text{ZnSO}}_4}{\text{1000}\text{mL}\text{solution}}\text{,}\frac{\text{1000}\text{mL}\text{solution}}{\text{0}\text{.500}\text{mol}{\text{ZnSO}}_4}$

Conclusion

The two pairs of unit factors for $0.500M{\text{ZnSO}}_{\text{4}}$ have been rightfully stated.