Chapter 13, Problem 57E

Interpretation Introduction

(a)

Interpretation:

The volume of solution containing $\text{2}\text{.50}\text{g}$ solute in $0.325M{\text{KNO}}_{\text{3}}$ 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 litre volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 57E

The volume of solution containing $\text{2}\text{.50}\text{g}$ solute in $0.325M{\text{KNO}}_{\text{3}}$ is $\text{0}\text{.0761}\text{L}$.

Explanation of Solution

It is given that the amount of solute is $\text{2}\text{.50}\text{g}$ and molarity of ${\text{KNO}}_{\text{3}}$ solution is $0.325M$.

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 oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of nitrogen is $14.01\text{g}{\text{mol}}^{-1}$

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

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{KNO}}_{\text{3}}& =& \text{Mass}\text{of}\text{K}+\text{mass}\text{of}\text{N}+\text{3}\times \text{mass}\text{of}\text{O}\\ & =& 39.10\text{g}{\text{mol}}^{-1}+14.01\text{g}{\text{mol}}^{-1}+\left(3\times 16.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 39.10\text{g}{\text{mol}}^{\text{-1}}+14.01\text{g}{\text{mol}}^{-1}+48.00\text{g}{\text{mol}}^{-1}\\ & =& 101.11\text{g}{\text{mol}}^{-1}\end{array}$

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 $\text{2}\text{.50}\text{g}$ and molar mass of solute as $101.11\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\begin{array}{rll}\text{n}& =& \frac{\text{2}\text{.5}\text{g}}{\text{101}\text{.11}\text{g}{\text{mol}}^{-1}}\\ & =& 0.0247\text{mol}\end{array}$

Substitute the values of number of moles as $\text{0}\text{.0247}\text{mol}$ and molarity of solution as $0.325M$ in equation (1) to calculate the volume of the solution as shown below.

$0.325=\frac{0.0247}{\text{V}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}\text{V}& =& \frac{0.0247}{0.325}\\ & =& 0.0761\text{L}\end{array}$

Therefore, the volume of ${\text{KNO}}_{\text{3}}$ solution is $\text{0}\text{.0761}\text{L}$.

Conclusion

The volume of ${\text{KNO}}_{\text{3}}$ solution containing $\text{2}\text{.50}\text{g}$ of ${\text{KNO}}_{\text{3}}$ is calculated as $\text{0}\text{.0761}\text{L}$.

Interpretation Introduction

(b)

Interpretation:

The volume of solution containing $\text{2}\text{.50}\text{g}$ solute in $0.325M{\text{AlBr}}_{\text{3}}$ 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 57E

The volume of solution containing $\text{2}\text{.50}\text{g}$ solute in $0.325M{\text{AlBr}}_{\text{3}}$ is $\text{0}\text{.0288}\text{L}$.

Explanation of Solution

It is given that the amount of solute is $\text{2}\text{.50}\text{g}$ and molarity of ${\text{AlBr}}_{\text{3}}$ solution is $0.325M$.

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 aluminum is $26.98\text{g}{\text{mol}}^{-1}$.

The molar mass of bromine is $79.90\text{g}{\text{mol}}^{-1}$.

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

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{AlBr}}_{\text{3}}& =& \text{Molar}\text{mass}\text{of}\text{Al}+\left(\text{3}\times \text{Molar}\text{mass}\text{of}\text{Br}\right)\\ & =& 26.98\text{g}{\text{mol}}^{-1}+\left(3\times 79.90\text{g}{\text{mol}}^{-1}\right)\\ & =& 26.98\text{g}{\text{mol}}^{\text{-1}}+239.7\text{g}{\text{mol}}^{-1}\\ & =& 266.68\text{g}{\text{mol}}^{-1}\end{array}$

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 $\text{2}\text{.50}\text{g}$ and molar mass of solute as $266.68\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\begin{array}{rll}\text{n}& =& \frac{\text{2}\text{.5}\text{g}}{266.68\text{g}{\text{mol}}^{-1}}\\ & =& 0.00937\text{mol}\end{array}$

Substitute the values of number of moles as $\text{0}\text{.00937}\text{mol}$ and molarity of solution as $0.325M$ in equation (1) to calculate the volume of the solution is shown below.

$0.325=\frac{0.00937}{\text{V}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}\text{V}& =& \frac{0.00937}{0.325}\\ & =& 0.0288\text{L}\end{array}$

Therefore, the volume of ${\text{AlBr}}_{\text{3}}$ solution is $\text{0}\text{.0288}\text{L}$.

Conclusion

The volume of ${\text{AlBr}}_{\text{3}}$ solution containing $\text{2}\text{.50}\text{g}$ of ${\text{AlBr}}_{\text{3}}$ is calculated as $\text{0}\text{.0288}\text{L}$.

Interpretation Introduction

(c)

Interpretation:

The volume of solution containing $\text{2}\text{.50}\text{g}$ solute in $1.00M\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ 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 litre volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 57E

The volume of solution containing $\text{2}\text{.50}\text{g}$ solute in $1.00M\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is $\text{0}\text{.0140}\text{L}$.

Explanation of Solution

It is given that the amount of solute is $\text{2}\text{.50}\text{g}$ and molarity of $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ solution is $1.00M$.

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 cobalt is $58.93\text{g}{\text{mol}}^{-1}$.

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

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

The molar mass of hydrogen is $1.01\text{g}{\text{mol}}^{\text{-1}}$.

Therefore, the molar mass of $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}& =& \text{Molar}\text{mass}\text{of}\text{Co}+\left(2\times \left(\begin{array}{c}\left(\text{2×Molar}\text{mass}\text{of}\text{C}\right)\\ +\left(\text{3×Molar}\text{mass}\text{of}\text{H}\right)\\ +\left(\text{2×Molar}\text{mass}\text{of}\text{O}\right)\end{array}\right)\right)\\ & =& \left(58.93\text{g}{\text{mol}}^{-1}+2\times \left(\begin{array}{l}\left(2\times 12.01\text{g}{\text{mol}}^{-1}\right)+\\ \left(3\times 1.01\text{g}{\text{mol}}^{-1}\right)+\left(2\times 16.00\text{g}{\text{mol}}^{-1}\right)\end{array}\right)\right)\\ & =& 58.93\text{g}{\text{mol}}^{-1}+\left(2\times 59.05\text{g}{\text{mol}}^{-1}\right)\\ & =& 59.93\text{g}{\text{mol}}^{-1}+118.1\text{g}{\text{mol}}^{-1}\end{array}$

The above equation gives the molar mass of $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ to be $178.03\text{g}{\text{mol}}^{-1}$

The formula to calculate the number of moles is given 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 $\text{2}\text{.50}\text{g}$ and molar mass of solute as $178.03\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\begin{array}{rll}\text{n}& =& \frac{\text{2}\text{.5}\text{g}}{178.03\text{g}{\text{mol}}^{-1}}\\ & =& 0.0140\text{mol}\end{array}$

Substitute the values of number of moles as $\text{0}\text{.0140}\text{mol}$ and molarity of solution as $1.00M$ in equation (1) to calculate the volume of the solution is shown below.

$1.00=\frac{0.0140}{\text{V}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}\text{V}& =& \frac{0.0140}{1.00}\\ & =& 0.0140\text{L}\end{array}$

Therefore, the volume of $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ solution is $\text{0}\text{.0140}\text{L}$.

Conclusion

The volume of $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ solution containing $\text{2}\text{.50}\text{g}$ of $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is calculated as $\text{0}\text{.0140}\text{L}$.

Interpretation Introduction

(d)

Interpretation:

The volume of solution containing $\text{2}\text{.50}\text{g}$ solute in $1.00M{\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}$ 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 57E

The volume of solution containing $\text{2}\text{.50}\text{g}$ solute in $1.00M{\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}$ is $\text{0}\text{.0168}\text{L}$.

Explanation of Solution

It is given that the amount of solute is $\text{2}\text{.50}\text{g}$ and molarity of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $1.00M$.

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 nitrogen is $14.01\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $1.01\text{g}{\text{mol}}^{-1}$.

The molar mass of phosphorus is $39.07\text{g}{\text{mol}}^{-1}$

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

Therefore, the molar mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}& =& \left(\begin{array}{c}\left(3\times \left(\text{Molar}\text{mass}\text{of}\text{N}\right)+\left(\text{4}\times \text{Molar}\text{mass}\text{of}\text{H}\right)\right)\\ +\text{Molar}\text{mass}\text{of}\text{P}+\left(4\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}\right)\\ & =& \left(\begin{array}{c}\left(3\times \left(14.01\text{g}{\text{mol}}^{-1}+4\times 1.01\right)\right)+30.97\text{g}{\text{mol}}^{-1}\\ +\left(4\times 16.00\text{g}{\text{mol}}^{-1}\right)\end{array}\right)\\ & =& 3\times 18.05\text{g}{\text{mol}}^{-1}+30.97\text{g}{\text{mol}}^{-1}+64.00\text{g}{\text{mol}}^{-1}\\ & =& 149.12\text{g}{\text{mol}}^{-1}\end{array}$

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 $\text{2}\text{.50}\text{g}$ and molar mass of solute as $149.12\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\begin{array}{rll}\text{n}& =& \frac{\text{2}\text{.5}\text{g}}{149.12\text{g}{\text{mol}}^{-1}}\\ & =& 0.01676\text{mol}\end{array}$

Substitute the values of number of moles as $\text{0}\text{.01676}\text{mol}$ and molarity of solution as $1.00M$ in equation (1) to calculate the volume of the solution is shown below.

$1.00=\frac{0.01676}{\text{V}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}\text{V}& =& \frac{0.0167}{1.00}\\ & =& 0.01676\text{L}\\ & \approx & \text{0}\text{.0168}\text{L}\end{array}$

Therefore, the volume of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $\text{0}\text{.0168}\text{L}$.

Conclusion

The volume of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}$ solution containing $\text{2}\text{.50}\text{g}$ of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}{\text{PO}}_{\text{4}}$ is calculated as $\text{0}\text{.0168}\text{L}$.