Chapter 8, Problem 8.91P

Interpretation Introduction

(a)

Interpretation:

The volume of $\text{2}\text{.5 M NaCl}$ solution required to prepare $\text{25 mL}$ of $\text{1 M}$ solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

  $\text{Number of moles}=\text{Molarity}\times \text{Volume}$

This can be rewritten as follows:

  $\text{MV}=\text{constant}$

Thus, the dilution formula is given as follows:

  ${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Answer to Problem 8.91P

The volume of $\text{2}\text{.5 M NaCl}$ solution required to prepare $\text{25 mL}$ of $\text{1 M}$ solution is $\text{10 mL}$.

Explanation of Solution

The dilution formula is given as follows:

  ${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Where,

• ${\text{M}}_1$denotes the initial molarity before dilution.
• ${\text{V}}_1$denotes the initial volume of solution.
• ${\text{M}}_2$denotes the initial molarity before dilution.
• ${\text{V}}_2$ denotes the final volume of solution.

The initial volume is given as $\text{25 mL}$.

Initial molarity is given as $\text{1 M}$.

The final volume after dilution is to be calculated.

The final molarity is $\text{2}\text{.5 M}$.

Substitute the values in the above formula to calculate the final molarity.

The dilution formula is given as follows:

  $\left(\text{1 M}\right)\left(\text{25 mL}\right)=\left(\text{2}\text{.5 M}\right){\text{V}}_2$

Rearrange the above expression to calculate the molarity.

  $\begin{array}{c}{\text{V}}_2=\frac{\left(\text{1 M}\right)\left(\text{25 mL}\right)}{2.5\text{ M}}\\ =\text{10 mL}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The volume of $\text{2}\text{.5 M NaCl}$ solution required to prepare $\text{1}\text{.5 L}$ of $\text{0}\text{.75 M}$ solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

  $\text{Number of moles}=\text{Molarity}\times \text{Volume}$

This can be rewritten as follows:

  $\text{MV}=\text{constant}$

Thus, the dilution formula is given as follows:

  ${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Answer to Problem 8.91P

The volume of $\text{2}\text{.5 M NaCl}$ solution required to prepare $\text{1}\text{.5 L}$ of $\text{0}\text{.75 M}$ solution is $450\text{ mL}$.

Explanation of Solution

The dilution formula is given as follows:

  ${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Where,

• ${\text{M}}_1$denotes the initial molarity before dilution.
• ${\text{V}}_1$denotes the initial volume of solution.
• ${\text{M}}_2$denotes the initial molarity before dilution.
• ${\text{V}}_2$ denotes the final volume of solution.

The initial volume is given as $\text{1}\text{.5 L}$.

Initial molarity is given as $\text{0}\text{.75 M}$.

The final volume after dilution is to be calculated.

The final molarity is $\text{2}\text{.5 M}$.

Substitute the values in the above formula to calculate the final molarity.

The dilution formula is given as follows:

  $\left(\text{0}\text{.75 M}\right)\left(\text{1}\text{.5 L}\right)=\left(\text{2}\text{.5 M}\right){\text{V}}_2$

Rearrange the above expression to calculate the molarity.

  $\begin{array}{c}{\text{V}}_2=\frac{\left(\text{0}\text{.75 M}\right)\left(\text{1}\text{.5 L}\right)}{2.5\text{ M}}\\ =0.45\text{ L}\end{array}$

The conversion factor to convert liters to milliliters is as follows:

  $1\text{}\text{L}=\text{1000 mL}$

Hence, convert the volume in liters to milliliters as follows:

  $\begin{array}{c}\text{Volume}\left(\text{mL}\right)=0.45\text{ L}\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =450\text{ mL }\end{array}$

Interpretation Introduction

(c)

Interpretation:

The volume of $\text{2}\text{.5 M NaCl}$ solution required to prepare $15\text{ mL}$ of $\text{0}\text{.25 M}$ solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

  $\text{Number of moles}=\text{Molarity}\times \text{Volume}$

This can be rewritten as follows:

  $\text{MV}=\text{constant}$

Thus, the dilution formula is given as follows:

  ${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Answer to Problem 8.91P

The volume of $\text{2}\text{.5 M NaCl}$ solution required to prepare $15\text{ mL}$ of $\text{0}\text{.25 M}$ solution is $1.5\text{ mL}$.

Explanation of Solution

The dilution formula is given as follows:

  ${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Where,

• ${\text{M}}_1$denotes the initial molarity before dilution.
• ${\text{V}}_1$denotes the initial volume of solution.
• ${\text{M}}_2$denotes the initial molarity before dilution.
• ${\text{V}}_2$ denotes the final volume of the solution.

The initial volume is given as $\text{15 mL}$.

Initial molarity is given as $\text{0}\text{.25 M}$.

The final volume after dilution is to be calculated.

The final molarity is $\text{2}\text{.5 M}$.

Substitute the values in the above formula to calculate the final molarity.

The dilution formula is given as follows:

  $\left(\text{0}\text{.25 M}\right)\left(\text{15 mL}\right)=\left(\text{2}\text{.5 M}\right){\text{V}}_2$

Rearrange the above expression to calculate the volume.

  $\begin{array}{c}{\text{V}}_2=\frac{\left(\text{0}\text{.25 M}\right)\left(\text{15 mL}\right)}{2.5\text{ M}}\\ =1.5\text{ mL}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The volume of $\text{2}\text{.5 M NaCl}$ solution required to prepare $\text{250 mL}$ of $\text{0}\text{.025 M}$ solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

  $\text{Number of moles}=\text{Molarity}\times \text{Volume}$

This can be rewritten as follows:

  $\text{MV}=\text{constant}$

Thus, the dilution formula is given as follows:

  ${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Answer to Problem 8.91P

The volume of $\text{2}\text{.5 M}\text{NaCl}$ solution required to prepare $\text{250 mL}$ of $\text{0}\text{.025 M}$ solution is $2.22\text{ mL}$.

Explanation of Solution

The dilution formula is given as follows:

  ${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Where,

• ${\text{M}}_1$denotes the initial molarity before dilution.
• ${\text{V}}_1$denotes the initial volume of the solution.
• ${\text{M}}_2$denotes the initial molarity before dilution.
• ${\text{V}}_2$ denotes the final volume of solution.

Initial volume is given as $\text{250 mL}$.

Initial molarity is given as $\text{0}\text{.025 M}$.

Final volume after dilution is to be calculated.

Final molarity is $\text{2}\text{.5 M}$.

Substitute the values in above formula to calculate the final molarity.

The dilution formula is given as follows:

  $\left(\text{0}\text{.025 M}\right)\left(\text{250 mL}\right)=\left(\text{2}\text{.5 M}\right){\text{V}}_2$

Rearrange the above expression to calculate the volume.

  $\begin{array}{c}{\text{V}}_2=\frac{\left(\text{0}\text{.25 M}\right)\left(\text{15 mL}\right)}{2.5\text{ M}}\\ =2.22\text{ mL}\end{array}$