Chapter 9.5, Problem 9PPC

Interpretation Introduction

Interpretation:

The required volume of stock solution to prepare the given volumes and concentrated final solutions are should be calculated by using given representation diagram.

Concept introduction:

Dilution:

• Dilution is the process, which is used to prepare a low concentrated solution from high concentrated solution.

Volumetric principle:

• In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation and it is,

$\begin{array}{l}{\text{M}}_{\text{c}}{\text{×mL}}_{\text{c}}\text{=}{\text{M}}_{\text{d}}{\text{×mL}}_{\text{d}}\\ {\text{M}}_{\text{c}}\text{=}\text{Initial}\text{concentration}\\ {\text{mL}}_{\text{c}}\text{=}\text{Initial}\text{volume}\\ {\text{M}}_{\text{d}}\text{=}\text{Final}\text{concentration}\\ {\text{mL}}_{\text{d}}\text{=}\text{Final}\text{volume}\end{array}$

Explanation of Solution

To calculate the required volumes of stock solutions to prepare the final diluted solutions

The given diagram is,

Diagram.1

To calculate the required volume of stock solution to prepare $\text{50 mL}$ of final solution.

$\begin{array}{l}InitialVolume=\frac{Finalnumberofsolute\times Finalvolume}{Initialnumberofsolute}\\ =\frac{3\times 50mL}{24}\\ =6.25mL\\ \end{array}$

From the given diagram (Diagram.1) the stock solution contains 24 numbers of solute particles and diluted solution contains 3 numbers of solute particles.

Consider a unit of stock solution contains 24 numbers of solute particles and a unit of diluted solution contains 3 numbers of solute particles.

The numbers of solute particles and final volume are plugged in the above equation to give required volume of stock solution to prepare $50mL$ of final solution.

The required volume of stock solution to prepare $50mL$ of final solution is $6.25mL$

The required volume of stock solution to prepare $100mL$ of final solution.

$\begin{array}{l}InitialVolume=\frac{Finalnumberofsolute\times Finalvolume}{Initialnumberofsolute}\\ =\frac{3\times 100mL}{24}\\ =12.5mL\\ \end{array}$

From the given diagram (Diagram.1) the stock solution contains 24 numbers of solute particles and diluted solution contains 3 numbers of solute particles.

Consider a unit of stock solution contains 24 numbers of solute particles and a unit of diluted solution contains 3 numbers of solute particles.

The numbers of solute particles and final volume are plugged in the above equation to give required volume of stock solution to prepare $100mL$ of final solution.

The required volume of stock solution to prepare $100mL$ of final solution is $12.5mL$

The required volume of stock solution to prepare $250mL$ of final solution.

$\begin{array}{l}InitialVolume=\frac{Finalnumberofsolute\times Finalvolume}{Initialnumberofsolute}\\ =\frac{3\times 250mL}{24}\\ =31.25mL\end{array}$

From the given diagram (Diagram.1) the stock solution contains 24 numbers of solute particles and diluted solution contains 3 numbers of solute particles.

Consider a unit of stock solution contains 24 numbers of solute particles and a unit of diluted solution contains 3 numbers of solute particles.

The numbers of solute particles and final volume are plugged in the above equation to give required volume of stock solution to prepare $250mL$ of final solution.

The required volume of stock solution to prepare $250mL$ of final solution is $31.25mL$.

Conclusion

The required volume of stock solution to prepare the given volumes and concentrated final solutions are calculated by using given representation diagram.