Chapter 13, Problem 58E

Interpretation Introduction

(a)

Interpretation:

The volume of solution that contains $\text{10}\text{.0}\text{g}$ solute in $0.275M\text{NaF}$ is to be calculated.

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 liter volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 58E

The volume of solution that contains $\text{10}\text{.0}\text{g}$ solute in $0.275M\text{NaF}$ is $\text{0}\text{.866}\text{L}$.

Explanation of Solution

It is given that the amount of solute is $\text{10}\text{.0}\text{g}$ and molarity of $\text{NaF}$ solution is $0.275M$.

The formula to determine molarity is shown below.

$M=\frac{\text{n}}{\text{V}}$ …(1)

Where

• $M$ is the molarity of a solution.

• $\text{n}$ is the number of moles of solute.

• $\text{V}$ is the volume of the solution.

The molar mass of sodium is $22.99\text{g}{\text{mol}}^{-1}$.

The molar mass of fluorine is $19.00\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of $\text{NaF}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}\text{NaF}& =& \text{Molar}\text{mass}\text{of}\text{Na}+\text{Molar}\text{mass}\text{of}\text{F}\\ & =& 22.99\text{g}{\text{mol}}^{-1}+19.00\text{g}{\text{mol}}^{-1}\\ & =& 41.99\text{g}{\text{mol}}^{-\text{1}}\end{array}$

The formula to calculate the number of moles is shown below.

$\text{n}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$ …(2)

Substitute the value of given mass of solute as $\text{10}\text{.0}\text{g}$ and molar mass of solute as $41.99\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\begin{array}{rll}\text{n}& =& \frac{\text{10}\text{.0}\text{g}}{41.99\text{g}{\text{mol}}^{-1}}\\ & =& 0.2381\text{mol}\end{array}$

Substitute the values of number of moles as $\text{0}\text{.2381}\text{mol}$ and molarity of solution as $0.275M$ in equation (1) to calculate the volume of the solution as shown below.

$0.275=\frac{0.2381}{\text{V}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}\text{V}& =& \frac{0.2381}{0.275}\\ & =& 0.8658\text{L}\\ \simeq \text{0}\text{.866}\text{L}\end{array}$

Therefore, the volume of $\text{NaF}$ solution is $\text{0}\text{.866}\text{L}$.

Conclusion

The volume of $\text{NaF}$ solution containing $\text{10}\text{.0}\text{g}$ $\text{NaF}$ is calculated as $\text{0}\text{.866}\text{L}$.

Interpretation Introduction

(b)

Interpretation:

The volume of solution that contains $\text{10}\text{.0}\text{g}$ solute in $0.275M{\text{CdCl}}_{\text{2}}$ is to be calculated.

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 liter volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 58E

The volume of solution that contains $\text{10}\text{.0}\text{g}$ solute in $0.275M{\text{CdCl}}_{\text{2}}$ is $0.198\text{L}$.

Explanation of Solution

It is given that the amount of solute is $\text{10}\text{.0}\text{g}$ and molarity of a solution of ${\text{CdCl}}_{\text{2}}$ is $0.275M$.

The formula to determine molarity is shown below.

$M=\frac{\text{n}}{\text{V}}$ …(1)

Where

• $M$ is the molarity of a solution.

• $\text{n}$ is the number of moles of solute.

• $\text{V}$ is the volume of the solution.

The molar mass of cadmium is $112.41\text{g}{\text{mol}}^{-1}$.

The molar mass of chlorine is $35.45\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{CdCl}}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{CdCl}}_{\text{2}}& =& \text{Molar}\text{mass}\text{of}\text{Cd}+\left(\text{2×Molar}\text{mass}\text{of}\text{Cl}\right)\\ & =& 112.41\text{g}{\text{mol}}^{-1}+\left(2\times 35.45\text{g}{\text{mol}}^{-1}\right)\\ & =& 112.41\text{g}{\text{mol}}^{-\text{1}}+70.9\text{g}{\text{mol}}^{-1}\\ & =& 183.31\text{g}{\text{mol}}^{-1}\end{array}$

The formula to calculate the number of moles is shown below.

$\text{n}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$ …(2)

Substitute the value of given mass of solute as $\text{10}\text{.0}\text{g}$ and molar mass of solute as $\text{183}\text{.31}\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\begin{array}{rll}\text{n}& =& \frac{10.0\text{g}}{183.31\text{g}{\text{mol}}^{-1}}\\ & =& 0.05455\text{mol}\end{array}$

Substitute the values of number of moles as $\text{0}\text{.05455}\text{mol}$ and molarity of solution as $0.275M$ in equation (1) to calculate the volume of the solution as shown below.

$0.275=\frac{0.05455}{\text{V}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}\text{V}& =& \frac{0.05455}{0.275}\\ & =& 0.198\text{L}\end{array}$

Therefore, the volume of ${\text{CdCl}}_{\text{2}}$ solution is $0.198\text{L}$.

Conclusion

The volume of ${\text{CdCl}}_{\text{2}}$ solution containing $\text{10}\text{.0}\text{g}$ ${\text{CdCl}}_{\text{2}}$ is calculated as $0.198\text{L}$.

Interpretation Introduction

(c)

Interpretation:

The volume of solution that contains $\text{10}\text{.0}\text{g}$ solute in $0.408M{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is to be calculated.

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 liter volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 58E

The volume of solution that contains $\text{10}\text{.0}\text{g}$ solute in $0.408M{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $0.177\text{L}$.

Explanation of Solution

It is given that the amount of solute is $\text{10}\text{.0}\text{g}$ and molarity of a solution of $0.408M$ is $0.408M$.

The formula to determine molarity is shown below.

$M=\frac{\text{n}}{\text{V}}$ …(1)

Where

• $M$ is the molarity of a solution.

• $\text{n}$ is the number of moles of solute.

• $\text{V}$ is the volume of the solution.

The molar mass of potassium is $39.10\text{g}{\text{mol}}^{-1}$.

The molar mass of carbon is $12.01\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-\text{1}}$

Therefore, the molar mass of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}& =& \left(\begin{array}{c}\left(\text{2×Molar}\text{mass}\text{of}\text{K}\right)+\text{Molar}\text{mass}\text{of}\text{C}\\ +\left(3\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}\right)\\ =\left(2\times 39.10\text{g}{\text{mol}}^{-1}\right)+12.01\text{g}{\text{mol}}^{-1}+\left(3\times 16.00\text{g}{\text{mol}}^{-1}\right)\\ =78.2\text{g}{\text{mol}}^{-\text{1}}+12.01\text{g}{\text{mol}}^{-1}+48.00\text{g}{\text{mol}}^{-1}\\ =138.21\text{g}{\text{mol}}^{-1}\end{array}$ The formula to calculate the number of moles is shown below.

$\text{n}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$ …(2)

Substitute the value of given mass of solute as $\text{10}\text{.0}\text{g}$ and molar mass of solute as $138.21\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\begin{array}{rll}\text{n}& =& \frac{\text{10}\text{.0}\text{g}}{138.21\text{g}{\text{mol}}^{-1}}\\ & =& 0.07235\text{mol}\end{array}$

Substitute the values of number of moles as $\text{0}\text{.07235}\text{mol}$ and molarity of solution as $0.408M$ in equation (1) to calculate the volume of the solution as shown below.

$0.408=\frac{0.07235}{\text{V}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}\text{V}& =& \frac{0.07235}{0.408}\\ & =& 0.177\text{L}\end{array}$

Therefore, the volume of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution is $0.177\text{L}$.

Conclusion

The volume of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution containing $\text{10}\text{.0}\text{g}$ ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as $0.177\text{L}$.

Interpretation Introduction

(d)

Interpretation:

The volume of solution that contains $\text{10}\text{.0}\text{g}$ solute in $0.408M\text{Fe}{\left({\text{ClO}}_3\right)}_{\text{3}}$ is to be calculated.

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 liter volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 58E

The volume of solution that contains $\text{10}\text{.0}\text{g}$ solute in $0.408M\text{Fe}{\left({\text{ClO}}_3\right)}_{\text{3}}$ is $0.080\text{L}$.

Explanation of Solution

It is given that the amount of solute is $\text{10}\text{.0}\text{g}$ and molarity of a solution of $\text{Fe}{\left({\text{ClO}}_3\right)}_{\text{3}}$ is $0.408M$.

The formula to determine molarity is shown below.

$M=\frac{\text{n}}{\text{V}}$ …(1)

Where

• $M$ is the molarity of a solution.

• $\text{n}$ is the number of moles of solute.

• $\text{V}$ is the volume of the solution.

The molar mass of iron is $55.85\text{g}{\text{mol}}^{-1}$.

The molar mass of chlorine is $35.45\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-\text{1}}$

Therefore, the molar mass of $\text{Fe}{\left({\text{ClO}}_3\right)}_{\text{3}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}\text{Fe}{\left({\text{ClO}}_3\right)}_{\text{3}}& =& \text{Molar}\text{mass}\text{of}\text{Fe}+\left(\text{3}\times \left(\begin{array}{l}\text{Molar}\text{mass}\text{of}\text{Cl}+\\ \left(\text{3}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}\right)\right)\\ =55.85\text{g}{\text{mol}}^{-1}+\left(3\times \left(35.45\text{g}{\text{mol}}^{-1}+3\times 16.00\text{g}{\text{mol}}^{-1}\right)\right)\\ =55.85\text{g}{\text{mol}}^{-1}+250.35\text{g}{\text{mol}}^{-1}\\ =306.2\text{g}{\text{mol}}^{-1}\end{array}$

The formula to calculate the number of moles is shown below.

$\text{n}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$ …(2)

Substitute the value of given mass of solute as $\text{10}\text{.0}\text{g}$ and molar mass of solute as $306.2\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\begin{array}{rll}\text{n}& =& \frac{10.0\text{g}}{306.2\text{g}{\text{mol}}^{-1}}\\ & =& 0.03265\text{mol}\end{array}$

Substitute the values of number of moles as $\text{0}\text{.03265}\text{mol}$ and molarity of solution as $0.408M$ in equation (1) to calculate the volume of the solution as shown below.

$0.408=\frac{0.03265}{\text{V}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}\text{V}& =& \frac{0.03265}{0.408}\\ & =& 0.080\text{L}\end{array}$

Therefore, the volume of $\text{Fe}{\left({\text{ClO}}_3\right)}_{\text{3}}$ solution is $0.080\text{L}$.

Conclusion

The volume of $\text{Fe}{\left({\text{ClO}}_3\right)}_{\text{3}}$ solution containing $\text{10}\text{.0}\text{g}$ $\text{Fe}{\left({\text{ClO}}_3\right)}_{\text{3}}$ is calculated as $0.080\text{L}$.