Chapter 13, Problem 45E

Interpretation Introduction

(a)

Interpretation:

The three pairs of unit factors for the aqueous solution having $\text{1}\text{.50\% KBr}$ mass/mass percent concentration are to be stated.

Concept introduction:

The percentage composition of a compound is found by comparing the mass contributed by each element to the molar mass of the substance. The sum of the percentage composition of each element is equal to the percentage composition of the compound.

Answer to Problem 45E

The first pair of unit factor is shown below.

$\frac{\text{1}\text{.50}\text{g}\text{KBr}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{1}\text{.50}\text{g}\text{KBr}}$

The second pair of unit factor is shown below.

$\frac{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{\text{1}\text{.50}\text{g}\text{KBr}}{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{1}\text{.50}\text{g}\text{KBr}}$

Explanation of Solution

The mass percentage of aqueous solution is given as $\text{1}\text{.50\% KBr}$. Therefore, the mass of $\text{KBr}$ is $\text{1}\text{.50}\text{g}$. Total mass of the solution is equal to $\text{100}\text{g}$. The first pair of unit factor is shown below.

$\frac{\text{1}\text{.50}\text{g}\text{KBr}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{1}\text{.50}\text{g}\text{KBr}}$

The mass of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ is calculated as shown below.

$\begin{array}{rll}\text{Mass}\text{of}\text{water}& =& \text{Mass}\text{of}\text{solution}-\text{Mass}\text{of}\text{KBr}\\ & =& \text{100}\text{g}-\text{1}\text{.50}\text{g}\\ & =& \text{98}\text{.50}\text{g}\end{array}$

The second pair of unit factor is shown below.

$\frac{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{\text{1}\text{.50}\text{g}\text{KBr}}{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{1}\text{.50}\text{g}\text{KBr}}$

Conclusion

The first pair of unit factor is shown below.

$\frac{\text{1}\text{.50}\text{g}\text{KBr}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{1}\text{.50}\text{g}\text{KBr}}$

The second pair of unit factor is shown below.

$\frac{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{\text{1}\text{.50}\text{g}\text{KBr}}{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{98}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{1}\text{.50}\text{g}\text{KBr}}$

Interpretation Introduction

(b)

Interpretation:

The three pairs of unit factors for the aqueous solution having $\text{2}{\text{.50\% A1C1}}_{\text{3}}$ mass/mass percent concentration are to be stated.

Concept introduction:

The percentage composition of a compound is found by comparing the mass contributed by each element to the molar mass of the substance. The sum of the percentage composition of each element is equal to the percentage composition of the compound.

Answer to Problem 45E

The first pair of unit factor is shown below.

$\frac{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}$

The second pair of unit factor is shown below.

$\frac{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}$

Explanation of Solution

The mass percentage of aqueous solution is given as $\text{2}{\text{.50\% A1C1}}_{\text{3}}$. Therefore, the mass of ${\text{A1C1}}_{\text{3}}$ is $\text{2}\text{.50}\text{g}$. Total mass of the solution is equal to $\text{100}\text{g}$. The first pair of unit factor is shown below.

$\frac{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}$

The mass of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ is calculated as shown below.

$\begin{array}{rll}\text{Mass}\text{of}\text{water}& =& \text{Mass}\text{of}\text{solution}-\text{Mass}\text{of}{\text{AlCl}}_3\\ & =& \text{100}\text{g}-2.\text{50}\text{g}\\ & =& \text{97}\text{.50}\text{g}\end{array}$

The second pair of unit factor is shown below.

$\frac{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}$

Conclusion

The first pair of unit factor is shown below.

$\frac{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}$

The second pair of unit factor is shown below.

$\frac{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{97}\text{.50}\text{g}{\text{H}}_2\text{O}}{\text{2}\text{.50}\text{g}{\text{A1C1}}_{\text{3}}}$

Interpretation Introduction

(c)

Interpretation:

The three pairs of unit factors for the aqueous solution having $\text{3}{\text{.75\% AgNO}}_{\text{3}}$ mass/mass percent concentration are to be stated.

Concept introduction:

The percentage composition of a compound is found by comparing the mass contributed by each element to the molar mass of the substance. The sum of the percentage composition of each element is equal to the percentage composition of the compound.

Answer to Problem 45E

The first pair of unit factor is shown below.

$\frac{3.75\text{g}{\text{AgNO}}_{\text{3}}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{3.75\text{g}{\text{AgNO}}_{\text{3}}}$

The second pair of unit factor is shown below.

$\frac{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{3.75\text{g}{\text{AgNO}}_{\text{3}}}{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}{3.75\text{g}{\text{AgNO}}_{\text{3}}}$

Explanation of Solution

The mass percentage of aqueous solution is given as $\text{3}{\text{.75\% AgNO}}_{\text{3}}$. Therefore, the mass of ${\text{AgNO}}_{\text{3}}$ is $\text{3}\text{.75}\text{g}$. Total mass of the solution is equal to $\text{100}\text{g}$. The first pair of unit factor is shown below.

$\frac{3.75\text{g}{\text{AgNO}}_{\text{3}}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{3.75\text{g}{\text{AgNO}}_{\text{3}}}$

The mass of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ is calculated as shown below.

$\begin{array}{rll}\text{Mass}\text{of}\text{water}& =& \text{Mass}\text{of}\text{solution}-\text{Mass}\text{of}{\text{AgNO}}_{\text{3}}\\ & =& \text{100}\text{g}-3.75\text{g}\\ & =& \text{96}\text{.25}\text{g}\end{array}$

The second pair of unit factor is shown below.

$\frac{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{3.75\text{g}{\text{AgNO}}_{\text{3}}}{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}{3.75\text{g}{\text{AgNO}}_{\text{3}}}$

Conclusion

The first pair of unit factor is shown below.

$\frac{3.75\text{g}{\text{AgNO}}_{\text{3}}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{3.75\text{g}{\text{AgNO}}_{\text{3}}}$

The second pair of unit factor is shown below.

$\frac{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{3.75\text{g}{\text{AgNO}}_{\text{3}}}{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{96}\text{.25}\text{g}{\text{H}}_2\text{O}}{3.75\text{g}{\text{AgNO}}_{\text{3}}}$

Interpretation Introduction

(d)

Interpretation:

The three pairs of unit factors for the aqueous solution having $\text{4}{\text{.25\% Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ mass/mass percent concentration are to be stated.

Concept introduction:

The percentage composition of a compound is found by comparing the mass contributed by each element to the molar mass of the substance. The sum of the percentage composition of each element is equal to the percentage composition of the compound.

Answer to Problem 45E

The first pair of unit factor is shown below.

$\frac{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}$

The second pair of unit factor is shown below.

$\frac{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}$

Explanation of Solution

The mass percentage of aqueous solution is given as $\text{4}{\text{.25\% Li}}_{\text{2}}{\text{SO}}_{\text{4}}$. Therefore, the mass of ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{4}\text{.25}\text{g}$. Total mass of the solution is equal to $\text{100}\text{g}$. The first pair of unit factor is shown below.

$\frac{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}$

The mass of water $\left({\text{H}}_{\text{2}}\text{O}\right)$ is calculated as shown below.

$\begin{array}{rll}\text{Mass}\text{of}\text{water}& =& \text{Mass}\text{of}\text{solution}-\text{Mass}\text{of}{\text{Li}}_2{\text{SO}}_4\\ & =& \text{100}\text{g}-4.25\text{g}\\ & =& \text{95}\text{.75}\text{g}\end{array}$

The second pair of unit factor is shown below.

$\frac{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}$

Conclusion

The first pair of unit factor is shown below.

$\frac{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}$

The second pair of unit factor is shown below.

$\frac{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}{\text{100}\text{g}\text{solution}}$ and $\frac{\text{100}\text{g}\text{solution}}{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}$

The third pair of unit factor is shown below.

$\frac{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}$ and $\frac{\text{95}\text{.75}\text{g}{\text{H}}_2\text{O}}{\text{4}\text{.25}\text{g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}$