Chapter 8, Problem 8.66EP

(a)

Interpretation Introduction

Interpretation:

The molarity of the prepared diluted solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.66EP

The molarity of the prepared diluted solution is 0.875 M.

Explanation of Solution

Given information:

Initial volume of solution = 35.0 mL

Initial concentration = 1.25 M

Final volume of solution = 50.0 mL

Final concentration = ?

Apply dilution equation to calculate the molarity of the diluted solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \\ {\text{C}}_{\text{d}}\text{=}\text{final}\text{concentration}\text{=}{\text{C}}_{\text{s}}\text{×}\frac{{\text{V}}_{\text{s}}}{{\text{V}}_{\text{d}}}\\ \text{=}\text{1}\text{.25}\text{M}\text{×}\frac{\text{35}\text{.0}\text{mL}}{\text{50}\text{.0}\text{mL}}\text{=}\text{0}\text{.875}\text{M}\end{array}$

Therefore, the molarity of the prepared diluted solution is 0.875 M.

(b)

Interpretation Introduction

Interpretation:

The molarity of the prepared diluted solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.66EP

The molarity of the prepared diluted solution is 0.461 M.

Explanation of Solution

Given information:

Initial volume of solution = 35.0 mL

Initial concentration = 1.25 M

Final volume of solution = 95.0 mL

Final concentration = ?

Apply dilution equation to calculate the molarity of the diluted solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \\ {\text{C}}_{\text{d}}\text{=}\text{final}\text{concentration}\text{=}{\text{C}}_{\text{s}}\text{×}\frac{{\text{V}}_{\text{s}}}{{\text{V}}_{\text{d}}}\\ \text{=}\text{1}\text{.25}\text{M}\text{×}\frac{\text{35}\text{.0}\text{mL}}{\text{95}\text{.0}\text{mL}}\text{=}\text{0}\text{.461}\text{M}\end{array}$

Therefore, the molarity of the prepared diluted solution is 0.461 M.

(c)

Interpretation Introduction

Interpretation:

The molarity of the prepared diluted solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.66EP

The molarity of the prepared diluted solution is 0.0449 M.

Explanation of Solution

Given information:

Initial volume of solution = 35.0 mL

Initial concentration = 1.25 M

Final volume of solution = 975 mL

Final concentration = ?

Apply dilution equation to calculate the molarity of the diluted solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \\ {\text{C}}_{\text{d}}\text{=}\text{final}\text{concentration}\text{=}{\text{C}}_{\text{s}}\text{×}\frac{{\text{V}}_{\text{s}}}{{\text{V}}_{\text{d}}}\\ \text{=}\text{1}\text{.25}\text{M}\text{×}\frac{\text{35}\text{.0}\text{mL}}{\text{975}\text{mL}}\text{=}\text{0}\text{.0449}\text{M}\end{array}$

Therefore, the molarity of the prepared diluted solution is 0.0449 M.

(d)

Interpretation Introduction

Interpretation:

The molarity of the prepared diluted solution has to be determined.

Concept Introduction:

Dilution equation:

Dilution equation is given by,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{where,}\\ {\text{C}}_{\text{s}}\text{-}\text{concentration}\text{of}\text{stock}\text{solution}\\ {\text{V}}_{\text{s}}\text{-}\text{volume}\text{of}\text{stock}\text{solution}\\ {\text{C}}_{\text{d}}\text{-}\text{concentration}\text{of}\text{diluted}\text{solution}\\ {\text{V}}_{\text{d}}\text{-}\text{volume}\text{of}\text{diluted}\text{solution}\end{array}$

Answer to Problem 8.66EP

The molarity of the prepared diluted solution is 0.0122 M.

Explanation of Solution

Given information:

Initial volume of solution = 35.0 mL

Initial concentration = 1.25 M

Final volume of solution = 3600 mL

Final concentration = ?

Apply dilution equation to calculate the molarity of the diluted solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \\ {\text{C}}_{\text{d}}\text{=}\text{final}\text{concentration}\text{=}{\text{C}}_{\text{s}}\text{×}\frac{{\text{V}}_{\text{s}}}{{\text{V}}_{\text{d}}}\\ \text{=}\text{1}\text{.25}\text{M}\text{×}\frac{\text{35}\text{.0}\text{mL}}{\text{3600}\text{mL}}\text{=}\text{0}\text{.0122}\text{M}\end{array}$

Therefore, the molarity of the prepared diluted solution is 0.0122 M.