Chapter 4, Problem 4.12P

(a)

Interpretation Introduction

Interpretation:

The solution that has the highest molarity is to be determined.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  $\text{Molarity of solution}\left(M\right)=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$

Answer to Problem 4.12P

The solution in beaker B has the highest molarity.

Explanation of Solution

Consider the particles present in the beaker as moles of solute.

The formula to calculate the molarity of solution in beaker is as follows:

  $\text{Molarity of solution}\left(M\right)=\frac{\text{number of moles}}{\text{volume of solution}}$        (1)

Substitute $8\text{mol}$ for number of moles and $50.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker A.

  $\begin{array}{c}\text{Molarity of solution}=\frac{8\text{mol}}{50.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =160\text{mol/L}\end{array}$

Substitute $12\text{mol}$ for number of moles and $50.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker B.

  $\begin{array}{c}\text{Molarity of solution}=\frac{12\text{mol}}{50.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =60000\text{mol/L}\end{array}$

Substitute $4\text{mol}$ for number of moles and $50.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker C.

  $\begin{array}{c}\text{Molarity of solution}=\frac{4\text{mol}}{50.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =80\text{mol/L}\end{array}$

Substitute $4\text{mol}$ for number of moles and $50.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker D.

  $\begin{array}{c}\text{Molarity of solution}=\frac{4\text{mol}}{50.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =80\text{mol/L}\end{array}$

Substitute $2\text{mol}$ for number of moles and $25.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker E.

  $\begin{array}{c}\text{Molarity of solution}=\frac{2\text{mol}}{25.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =80\text{mol/L}\end{array}$

Substitute $4\text{mol}$ for number of moles and $25.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker F.

  $\begin{array}{c}\text{Molarity of solution}=\frac{4\text{mol}}{25.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =160\text{mol/L}\text{.}\end{array}$

Hence, the solution in beaker B has the highest molarity.

Conclusion

Molarity $\left(M\right)$ is directly proportional to number of moles and inversely related to the volume of solution.

(b)

Interpretation Introduction

Interpretation:

The solutions that have the same molarity are to be determined.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  $\text{Molarity of solution}\left(M\right)=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$

Answer to Problem 4.12P

The solution in beaker A and F has the same molarity and solution in beaker C, D and E have the same molarity.

Explanation of Solution

Consider the particles present in the beaker as moles of solute.

The formula to calculate the molarity of solution in beaker is as follows:

  $\text{Molarity of solution}\left(M\right)=\frac{\text{number of moles}}{\text{volume of solution}}$        (1)

Substitute $8\text{mol}$ for number of moles and $50.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker A.

  $\begin{array}{c}\text{Molarity of solution}=\frac{8\text{mol}}{50.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =160\text{mol/L}\end{array}$

Substitute $12\text{mol}$ for number of moles and $50.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker B.

  $\begin{array}{c}\text{Molarity of solution}=\frac{12\text{mol}}{50.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =60000\text{mol/L}\end{array}$

Substitute $4\text{mol}$ for number of moles and $50.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker C.

  $\begin{array}{c}\text{Molarity of solution}=\frac{4\text{mol}}{50.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =80\text{mol/L}\end{array}$

Substitute $4\text{mol}$ for number of moles and $50.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker D.

  $\begin{array}{c}\text{Molarity of solution}=\frac{4\text{mol}}{50.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =80\text{mol/L}\end{array}$

Substitute $2\text{mol}$ for number of moles and $25.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker E.

  $\begin{array}{c}\text{Molarity of solution}=\frac{2\text{mol}}{25.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =80\text{mol/L}\end{array}$

Substitute $4\text{mol}$ for number of moles and $25.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of solution in beaker F.

  $\begin{array}{c}\text{Molarity of solution}=\frac{4\text{mol}}{25.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =160\text{mol/L}\text{.}\end{array}$

The solution in beaker A and F has the same molarity and the value of molarity is $160\text{mol/L}$. The solution in beaker C, D and E has the same molarity and the value of molarity is $80\text{mol/L}$.

Conclusion

Molarity $\left(M\right)$ is directly proportional to number of moles and inversely related to the volume of solution.

(c)

Interpretation Introduction

Interpretation:

Whether the mixture of solution A and C have a higher, a lower, or the same molarity as solution B is to be determined.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  $\text{Molarity of solution}\left(M\right)=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$

Answer to Problem 4.12P

The mixture of solution A and C has a lower molarity as compared to solution B.

Explanation of Solution

The number of moles in solution A is $8\text{mol}$ and number of moles in solution C is $4\text{mol}$. Therefore the total number of moles in the mixture of A and C is $12\text{mol}$.

The formula to calculate the total volume is as follows:

  $\text{Total volume}=V_{\text{A}}+V_{\text{C}}$        (2)

Substitute $50.00\text{mL}$ for $V_{\text{A}}$ and $50.00\text{mL}$ for $V_{\text{C}}$ in the equation (2).

  $\begin{array}{c}\text{Total volume}=50.00\text{mL}+50.00\text{mL}\\ =100.00\text{mL}\end{array}$

Substitute $12\text{mol}$ for number of moles and $100.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of mixture of solution A and C.

  $\begin{array}{c}\text{Molarity of solution}=\frac{12\text{mol}}{100.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =120\text{mol/L}\text{.}\end{array}$

The mixture of solution A and C has lower molarity as compared to solution B.

Conclusion

Molarity $\left(M\right)$ is directly proportional to number of moles and inversely related to the volume of solution. The new moles and volume is calculated by the summation of the volumes and moles of the solution that are mixed.

(d)

Interpretation Introduction

Interpretation:

Whether the molarity when $50\text{ mL}$ of water is added to solution D is higher, lower, or the same as the molarity of solution F after $75\text{ mL}$ is added to it is to be determined.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  $\text{Molarity of solution}\left(M\right)=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$

Answer to Problem 4.12P

The molarity when $50\text{ mL}$ of water is added to solution D is same as the molarity of solution F after $75\text{ mL}$ is added to it.

Explanation of Solution

The volume of solution D is $50.00\text{mL}$. The new volume of solution D is calculated as follows:

  $\begin{array}{c}\text{New volume of}\text{solution D}=50.00\text{mL}+50.00\text{mL}\\ =100.00\text{mL}\end{array}$

The volume of solution F is $25.00\text{mL}$. The new volume of solution F is calculated as follows:

  $\begin{array}{c}\text{New volume of}\text{solution F}=25.00\text{mL}+75.00\text{mL}\\ =100.00\text{mL}\end{array}$

Substitute $4\text{mol}$ for number of moles and $100.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of new solution in beaker D.

  $\begin{array}{c}\text{Molarity of solution}=\frac{4\text{mol}}{100.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =40\text{mol/L}\end{array}$

Substitute $4\text{mol}$ for number of moles and $100.00\text{mL}$ for volume of solution in the equation (1) to calculate the molarity of new solution in beaker F.

  $\begin{array}{c}\text{Molarity of solution}=\frac{4\text{mol}}{100.00\text{mL}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =40\text{mol/L}\text{.}\end{array}$

The molarity of solution D is same as the molarity of solution F.

Conclusion

Molarity $\left(M\right)$ is directly proportional to number of moles and inversely related to the volume of solution.

(e)

Interpretation Introduction

Interpretation:

The solvent must be evaporated from solution E for it to have the same molarity as solution A is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the molarity of solution when moles of solute and volume of solution are given is as follows:

  $\text{Molarity of solution}\left(M\right)=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$

Answer to Problem 4.12P

$12.5\text{mL}$ of solvent must be evaporated from solution E for it to have the same molarity as solution A.

Explanation of Solution

The molarity of solution E should be equal to solution A. Therefore the molarity os solution should be $160\text{mol/L}$.

The formula to calculate the molarity of solution in beaker is as follows:

  $\text{Molarity of solution}\left(M\right)=\frac{\text{number of moles}}{\text{volume of solution}}$        (3)

Rearrange the equation (3) to calculate the volume of solution evaporated.

  $\text{Volume of solution}=\frac{\text{number of moles}}{\text{Molarity of solution}}$        (4)

Substitute $2\text{mol}$ for number of moles and $160\text{mol/L}$ for molarity of solution in the equation (4) to calculate the volume of solution.

$\begin{array}{c}\text{Volume of solution}=\frac{2\text{mol}}{160\text{mol/L}}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =12.5\text{mL}\text{.}\end{array}$

Therefore to get the same molarity as of the solution A, $2\text{mol}$ should be present in $12.5\text{mL}$. Hence, $12.5\text{mL}$ of solvent must be evaporated from solution E.

Conclusion

Molarity $\left(M\right)$ is directly proportional to number of moles and inversely related to the volume of solution.