Chapter 15, Problem 31A

(a)

Interpretation Introduction

Interpretation:

The new molarity needs to be determined if 55 mL of water is added to 25 mL of 0.119 M NaCl solution.

Concept Introduction:

• The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

Where, $M_1$ is the molarity in initial conditions while $M_2$ is the molarity of the solution in final conditions. Similarly, $V_1$ is the volume of the solution in initial conditions while $V_2$ is the volume in final conditions.

Answer to Problem 31A

  $\text{0}\text{.0541}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$

Explanation of Solution

Given information:

The volume is, $V_1=25.\text{0mL}\text{and}{V}_2=5\text{5mL}$

The molarity is, $M_1=0.119\text{M}$

Calculation:

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

By substituting the given values in the formula and getting the molarity after the dilution,

  $\begin{array}{l}\left(\text{0}\text{.119M}\right)\text{×}\left(\text{25}\text{.0mL}\right)=M_2\left(\text{55mL}\right)\\ M_2=\frac{\left(\text{0}\text{.119M}\right)\text{×}\left(\text{25}\text{.0mL}\right)}{\left(\text{55mL}\right)}\\ =\text{0}\text{.0541}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$

The molarity is $\text{0}\text{.0541}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$ .

(b)

Interpretation Introduction

Interpretation:

The new molarity needs to be determined if 125 mL of water is added to 45.3 mL of 0.701 M NaOH solution.

Concept Introduction:

• The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

Where, $M_1$ is the molarity in initial conditions while $M_2$ is the molarity of the solution in final conditions. Similarly, $V_1$ is the volume of the solution in initial conditions while $V_2$ is the volume in final conditions.

Answer to Problem 31A

  $0.25\text{4}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$

Explanation of Solution

Given information:

The volume is, $V_1=45.3mL\text{and}{V}_2=\text{125mL}$

The molarity is, $M_1=0.7\text{01M}$

Calculation:

The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

By substituting the given values in the formula and getting the molarity after the dilution,

  $\begin{array}{l}\left(\text{0}\text{.701M}\right)\text{×}\left(\text{45}\text{.3mL}\right)=M_2\left(\text{125mL}\right)\\ M_2=\frac{\left(\text{0}\text{.701M}\right)\text{×}\left(\text{45}\text{.3mL}\right)}{\left(\text{125mL}\right)}\\ =0.25\text{4}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$

The molarity is $\text{0}\text{.254}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$ .

(c)

Interpretation Introduction

Interpretation:

The new molarity of the solution needs to be determined if 550 mL of water is added to 125 mL of 3.01 M KOH solution.

Concept Introduction:

• The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

Where, $M_1$ is the molarity in initial conditions while $M_2$ is the molarity of the solution in final conditions. Similarly, $V_1$ is the volume of the solution in initial conditions while $V_2$ is the volume in final conditions.

Answer to Problem 31A

  $0.68\text{4}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$

Explanation of Solution

Given information:

The volume is, $V_1=12\text{5mL}\text{and}{V}_2=\text{550mL}$

The molarity is, $M_1=3.\text{01M}$

Calculation:

The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

By substituting the given values in the formula and getting the molarity after the dilution,

  $\begin{array}{l}\left(0.701\text{M}\right)\times \left(45.3\text{mL}\right)=M_2\left(\text{125mL}\right)\\ M_2=\frac{\left(\text{3}\text{.01M}\right)\text{×}\left(\text{125mL}\right)}{\left(\text{550mL}\right)}\\ \text{=}\text{0}\text{.684}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\end{array}$

The molarity is $0.6\text{84}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$ .

(d)

Interpretation Introduction

Interpretation:

The new molarity of the solution needs to be determined if 335 mL of water is added to 75.3 mL of 2.07 M CaCl₂ solution.

Concept Introduction:

• The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

• The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

Where, $M_1$ is the molarity in initial conditions while $M_2$ is the molarity of the solution in final conditions. Similarly, $V_1$ is the volume of the solution in initial conditions while $V_2$ is the volume in final conditions.

Answer to Problem 31A

  $\text{0}\text{.465}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$

Explanation of Solution

Given information:

The volume is, $V_1=75.3mLand{V}_2=335mL$

The molarity is, $M_1=2.\text{07M}$

Calculation:

The equation of dilution is given as,

  $M_1{V}_1=M_2{V}_2$

By substituting the given values in the formula and getting the molarity after the dilution,

  $\begin{array}{l}\left(\text{0}\text{.701M}\right)\text{×}\left(\text{45}\text{.3mL}\right)=M_2\left(\text{125mL}\right)\\ M_2=\frac{\left(\text{2}\text{.07M}\right)\text{×}\left(\text{75}\text{.3mL}\right)}{\left(\text{335mL}\right)}\\ =0.465\raisebox{1ex}{$mol$}\!\left/ \!\raisebox{-1ex}{$L$}\right.\end{array}$

The molarity is $0.46\text{5}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$ .