Chapter 8, Problem 8.67EP

(a)

Interpretation Introduction

Interpretation:

The volume of water should be added to yield the given 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 8.67EP

The volume of water should be added to yield the given solution is 1450 mL.

Explanation of Solution

Given information:

Volume of solution = 50.0 mL

Initial concentration = 3.00 M

Required concentration = 0.100 M

Final volume of solution = ?

Apply dilution equation to calculate the volume of water should be added to yield the given solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{rearrange}\text{the}\text{above}\text{equation,}\\ {\text{V}}_{\text{d}}\text{=}\text{diluted}\text{volume}\text{=}{\text{V}}_{\text{s}}\text{×}\frac{{\text{C}}_{\text{s}}}{{\text{C}}_{\text{d}}}\\ \text{=}\text{50}\text{.0}\text{mL}\text{×}\frac{\text{3}\text{.00}\text{M}}{\text{0}\text{.100}\text{M}}\text{=}\text{1500}\text{mL}\end{array}$

Now subtract the original volume (V_d) from the final volume (V_s)

$\begin{array}{l}\text{Volume}\text{of}\text{water}\text{added}\text{=}{\text{V}}_{\text{d}}\text{-}{\text{V}}_{\text{s}}\\ \text{=}\text{1500}\text{mL}\text{-}\text{50}\text{.0}\text{mL}\text{=}\text{1450}\text{mL}\end{array}$

Therefore, the volume of water should be added to yield the given solution is 1450 mL.

(b)

Interpretation Introduction

Interpretation:

The volume of water should be added to yield the given 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 8.67EP

The volume of water should be added to yield the given solution is 18.0 mL.

Explanation of Solution

Given information:

Volume of solution = 2.00 mL

Initial concentration = 1.00 M

Required concentration = 0.100 M

Final volume of solution = ?

Apply dilution equation to calculate the volume of water should be added to yield the given solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{rearrange}\text{the}\text{above}\text{equation,}\\ {\text{V}}_{\text{d}}\text{=}\text{diluted}\text{volume}\text{=}{\text{V}}_{\text{s}}\text{×}\frac{{\text{C}}_{\text{s}}}{{\text{C}}_{\text{d}}}\\ \text{=}\text{2}\text{.00}\text{mL}\text{×}\frac{\text{1}\text{.00}\text{M}}{\text{0}\text{.100}\text{M}}\text{=}\text{20}\text{.0}\text{mL}\end{array}$

Now subtract the original volume (V_d) from the final volume (V_s)

$\begin{array}{l}\text{Volume}\text{of}\text{water}\text{added}\text{=}{\text{V}}_{\text{d}}\text{-}{\text{V}}_{\text{s}}\\ \text{=}\text{20}\text{mL}\text{-}\text{2}\text{mL}\text{=}\text{18}\text{mL}\end{array}$

Therefore, the volume of water should be added to yield the given solution is 18.0 mL.

(c)

Interpretation Introduction

Interpretation:

The volume of water should be added to yield the given 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 8.67EP

The volume of water should be added to yield the given solution is 85550 mL.

Explanation of Solution

Given information:

Volume of solution = 1450 mL

Initial concentration = 6.00 M

Required concentration = 0.100 M

Final volume of solution = ?

Apply dilution equation to calculate the volume of water should be added to yield the given solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{rearrange}\text{the}\text{above}\text{equation,}\\ {\text{V}}_{\text{d}}\text{=}\text{diluted}\text{volume}\text{=}{\text{V}}_{\text{s}}\text{×}\frac{{\text{C}}_{\text{s}}}{{\text{C}}_{\text{d}}}\\ \text{=}\text{1450}\text{mL}\text{×}\frac{\text{6}\text{.00}\text{M}}{\text{0}\text{.100}\text{M}}\text{=}\text{87000}\text{mL}\end{array}$

Now subtract the original volume (V_d) from the final volume (V_s)

$\begin{array}{l}\text{Volume}\text{of}\text{water}\text{added}\text{=}{\text{V}}_{\text{d}}\text{-}{\text{V}}_{\text{s}}\\ \text{=}\text{87000}\text{mL}\text{-}\text{1450}\text{mL}\text{=}\text{85550}\text{mL}\end{array}$

Therefore, the volume of water should be added to yield the given solution is 85550 mL.

(d)

Interpretation Introduction

Interpretation:

The volume of water should be added to yield the given 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 8.67EP

The volume of water should be added to yield the given solution is 7.5 mL.

Explanation of Solution

Given information:

Volume of solution = 75.0 mL

Initial concentration = 0.110 M

Required concentration = 0.100 M

Final volume of solution = ?

Apply dilution equation to calculate the volume of water should be added to yield the given solution,

$\begin{array}{l}{\text{C}}_{\text{s}}\text{×}{\text{V}}_{\text{s}}\text{=}{\text{C}}_{\text{d}}\text{×}{\text{V}}_{\text{d}}\\ \text{rearrange}\text{the}\text{above}\text{equation,}\\ {\text{V}}_{\text{d}}\text{=}\text{diluted}\text{volume}\text{=}{\text{V}}_{\text{s}}\text{×}\frac{{\text{C}}_{\text{s}}}{{\text{C}}_{\text{d}}}\\ \text{=}\text{75}\text{.0}\text{mL}\text{×}\frac{\text{0}\text{.110}\text{M}}{\text{0}\text{.100}\text{M}}\text{=}\text{82}\text{.5}\text{mL}\end{array}$

Now subtract the original volume (V_d) from the final volume (V_s)

$\begin{array}{l}\text{Volume}\text{of}\text{water}\text{added}\text{=}{\text{V}}_{\text{d}}\text{-}{\text{V}}_{\text{s}}\\ \text{=}\text{82}\text{.5}\text{mL}\text{-}\text{75}\text{.0}\text{mL}\text{=}\text{7}\text{.5}\text{mL}\end{array}$

Therefore, the volume of water should be added to yield the given solution is 7.5 mL.