Chapter 6, Problem 6.35P

6-35 Describe how we would prepare the following solutions:

(a) 280. mL of a 27% v/v solution of ethanol C₂H₆O, in water

(b) 435 mL of a 1.8% v/v solution of ethyl acetate, C₄H₈O₂ in water

(c) 1.65 L of an 8.00% v/v solution of benzene C₆H₆, in chloroform, CHCI₃

Interpretation Introduction

Interpretation:

The formation of following solutions should be explained:

280.0 mL of a 27% v/v solution of ethanol, ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$ in water.

Concept Introduction:

Concentration has several ways to be calculated. It typically relates an amount of solute and the solution. In volume of solute per unit volume of solvent, or v/v, the total volume of the solute, solvent and solution must be known.

The formula for v/v is as follows:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solvent}}}\times 100\%$.

Here, $V_{\text{solute}}$ is volume of solute and $V_{\text{solvent}}$ is volume of solvent.

When solving for the required volume of solute, the following formula is used:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solute}}+V_{\text{solvent}}}\times 100\%$.

Answer to Problem 6.35P

$75.\text{6 mL}$ of the solute must be added.

Explanation of Solution

Given:

$\begin{array}{l}V\text{}=280.\text{0 mL}\\ \text{C}\text{}=\text{}27\%\text{v/v}\end{array}$.

The formula for v/v is as follows:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solvent}}}\times 100\%$.

Here, $V_{\text{solute}}$ is volume of solute and $V_{\text{solvent}}$ is volume of solvent.

When solving for the required volume of solute, the following formula is used:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solute}}+V_{\text{solvent}}}\times 100\%$.

Substitute known data and solve for mass of solute.

$\begin{array}{l}{\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solute}}+V_{\text{solvent}}}\times 100\%\\ 27\%v/v=\frac{V_{\text{solute}}}{28\text{0 mL}}\times 100\%\\ V_{solute}=\frac{27\%v/v}{100\%}(28\text{0 mL})\\ V_{solute}=75.\text{6 mL}\end{array}$

Thus, add $75.\text{6 mL}$ of solute and then add the remaining volume of water.

Interpretation Introduction

Interpretation:

The formation of following solutions should be explained:

435 mL of a 1.8 % v/v solution of ethyl acetate ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}$ in water.

Concept Introduction:

Concentration has several ways to be calculated. It typically relates an amount of solute and the solution. In volume of solute per unit volume of solvent, or v/v, the total volume of the solute, solvent and solution must be known.

The formula for v/v is as follows:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solvent}}}\times 100\%$.

Here, $V_{\text{solute}}$ is volume of solute and $V_{\text{solvent}}$ is volume of solvent.

When solving for the required volume of solute, the following formula is used:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solute}}+V_{\text{solvent}}}\times 100\%$.

Answer to Problem 6.35P

$7.83\text{m}\text{L}$ of the solute must be added.

Explanation of Solution

Given:

$\begin{array}{l}V\text{}=43\text{5mL}\\ \text{C}\text{}=\text{}1.8\%v/v\end{array}$.

The formula for v/v is as follows:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solvent}}}\times 100\%$.

Here, $V_{\text{solute}}$ is volume of solute and $V_{\text{solvent}}$ is volume of solvent.

When solving for the required volume of solute, the following formula is used:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solute}}+V_{\text{solvent}}}\times 100\%$.

Substitute known data and solve for mass of solute.

$\begin{array}{l}{\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solute}}+V_{\text{solvent}}}\text{X}100\%\\ 1.8\%v/v=\frac{V_{\text{solute}}}{43\text{5 mL}}\times 100\%\\ V_{solute}=\frac{1.8\%v/v}{100\%}\left(43\text{5mL}\right)\\ V_{solute}=7.8\text{3 mL}\end{array}$

Then, add $7.8\text{3 mL}$ volume of solute to the remaining volume of water.

Interpretation Introduction

Interpretation:

The formation of following solutions should be explained:

1.65 L of a 8.00 % v/v solution of benzene, ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ in chloroform, ${\text{CHCl}}_{\text{3}}$.

Concept Introduction:

Concentration has several ways to be calculated. It typically relates an amount of solute and the solution. In volume of solute per unit volume of solvent, or v/v, the total volume of the solute, solvent and solution must be known.

The formula for v/v is as follows:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solvent}}}\times 100\%$.

Here, $V_{\text{solute}}$ is volume of solute and $V_{\text{solvent}}$ is volume of solvent.

When solving for the required volume of solute, the following formula is used:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solute}}+V_{\text{solvent}}}\times 100\%$.

Answer to Problem 6.35P

$13\text{2 mL}$ of the solute must be added.

Explanation of Solution

Given:

$\begin{array}{l}V\text{}=1.65\text{L}=1650\text{m}\text{L}\\ \text{C}\text{}=\text{}8.00\%v/v\end{array}$.

The formula for v/v is as follows:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solvent}}}\times 100\%$.

Here, $V_{\text{solute}}$ is volume of solute and $V_{\text{solvent}}$ is volume of solvent.

When solving for the required volume of solute, the following formula is used:

${\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solute}}+V_{\text{solvent}}}\times 100\%$.

Substitute known data and solve for mass of solute.

$\begin{array}{l}{\text{C}}_{v/v}=\frac{V_{\text{solute}}}{V_{\text{solute}}+V_{\text{solvent}}}\times 100\%\\ 8.00\%v/v=\frac{V_{\text{solute}}}{1650\text{mL}}\times 100\%\\ V_{solute}=\frac{8.0\%v/v}{100\%}\left(165\text{0 mL}\right)\\ V_{solute}=13\text{2 mL}\end{array}$

Then, add $132\text{ mL}$ of solute is added to the remaining volume of water.