Chapter 7, Problem 7.35E

Calculate the concentration in $\%\left(\text{v}/\text{v}\right)$ of the following solutions:

a. $\text{250}\text{.}\text{mL}$ of solution contains $\text{15}\text{.0}\text{mL}$ of acetone.

b. $\text{250}\text{.}\text{mL}$ of solution contains $\text{15}\text{.0}\text{mL}$ of any soluble liquid solute.

c. $\text{1}\text{.0}$ quart of acetic acid is put into a $\text{5-}$ gallon container, and enough water is added to fill the container.

d.A solution of acetone and water is separated by distillation. A $\text{300}\text{.-mL}$ sample gives $109\text{mL}$ of acetone.

Interpretation Introduction

(a)

Interpretation:

The concentration in $\%\left(\text{v}/\text{v}\right)$ of $\text{250}\text{.}\text{mL}$ of solution that contains $\text{15}\text{.0}\text{mL}$ of acetone is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume is expressed in terms of percent weight by volume $\left(\text{\%}\text{w}/\text{v}\right)$, percent volume by volume $\left(\text{\%}\text{v}/\text{v}\right)$ and molarity $\left(\text{M}\right)$. When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight $\left(\text{\%}\text{w}/\text{w}\right)$.

Answer to Problem 7.35E

The concentration in $\%\left(\text{v}/\text{v}\right)$ of $\text{250}\text{.}\text{mL}$ of solution that contains $\text{15}\text{.0}\text{mL}$ of acetone is $6.0\%\text{v}/\text{v}$.

Explanation of Solution

The formula to calculate concentration in $\%\left(\text{v}/\text{v}\right)$ is given below as,

$\%\left(\text{v}/\text{v}\right)=\frac{\text{Solute}\text{volume}}{\text{Solution}\text{volume}}\times 100$

The volume of acetone and volume of solution is $\text{15}\text{.0}\text{mL}$ and $\text{250}\text{.}\text{mL}$ respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

$\begin{array}{rll}\%\left(\text{v}/\text{v}\right)& =& \frac{\text{volume}\text{of}\text{acetone}}{\text{Solution}\text{volume}}\times 100\\ & =& \frac{15\text{mL}}{250\text{mL}}\times 100\\ & =& 6.0\%\text{v}/\text{v}\end{array}$

Thus, the concentration in $\%\left(\text{v}/\text{v}\right)$ of $\text{250}\text{.}\text{mL}$ of solution that contains $\text{15}\text{.0}\text{mL}$ of acetone is $6.0\%\text{v}/\text{v}$.

Conclusion

The concentration in $\%\left(\text{v}/\text{v}\right)$ of $200.\text{mL}$ of solution containing $15\text{mL}$ of alcohol is $6.0\%\text{v}/\text{v}$.

Interpretation Introduction

(b)

Interpretation:

The concentration in $\%\left(\text{v}/\text{v}\right)$ of $\text{250}\text{.}\text{mL}$ of solution that contains $\text{15}\text{.0}\text{mL}$ of any soluble liquid solute is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume $\left(\text{\%}\text{w}/\text{v}\right)$, percent volume by volume $\left(\text{\%}\text{v}/\text{v}\right)$ and molarity $\left(\text{M}\right)$. When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight $\left(\text{\%}\text{w}/\text{w}\right)$.

Answer to Problem 7.35E

The concentration in $\%\left(\text{v}/\text{v}\right)$ of $\text{250}\text{.}\text{mL}$ of solution that contains $\text{15}\text{.0}\text{mL}$ of any soluble liquid solute is $6.0\%\text{v}/\text{v}$.

Explanation of Solution

The formula to calculate concentration in $\%\left(\text{v}/\text{v}\right)$ is given below as,

$\%\left(\text{v}/\text{v}\right)=\frac{\text{Solute}\text{volume}}{\text{Solution}\text{volume}}\times 100$

The volume of solute and volume of solution is $\text{15}\text{.0}\text{mL}$ and $\text{250}\text{.}\text{mL}$ respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

$\begin{array}{rll}\%\left(\text{v}/\text{v}\right)& =& \frac{\text{volume}\text{of}\text{solute}}{\text{Solution}\text{volume}}\times 100\\ & =& \frac{15\text{mL}}{250\text{mL}}\times 100\\ & =& 6.0\%\text{v}/\text{v}\end{array}$

Thus, the concentration in $\%\left(\text{v}/\text{v}\right)$ of $\text{250}\text{.}\text{mL}$ of solution that contains $\text{15}\text{.0}\text{mL}$ of any soluble liquid solute is $6.0\%\text{v}/\text{v}$.

Conclusion

The concentration in $\%\left(\text{v}/\text{v}\right)$ of $\text{250}\text{.}\text{mL}$ of solution that contains $\text{15}\text{.0}\text{mL}$ of any soluble liquid solute is $6.0\%\text{v}/\text{v}$.

Interpretation Introduction

(c)

Interpretation:

The concentration in $\%\left(\text{v}/\text{v}\right)$ of $\text{1}\text{.0}$ quart of acetic acid that is put into a $\text{5-}$ gallon container is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume $\left(\text{\%}\text{w}/\text{v}\right)$, percent volume by volume $\left(\text{\%}\text{v}/\text{v}\right)$ and molarity $\left(\text{M}\right)$. When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight $\left(\text{\%}\text{w}/\text{w}\right)$.

Answer to Problem 7.35E

The concentration in $\%\left(\text{v}/\text{v}\right)$ of $\text{1}\text{.0}$ quart of acetic acid that is put into a $\text{5-}$ gallon container is $5\%\left(\text{v}/\text{v}\right)$.

Explanation of Solution

The formula to calculate concentration in $\%\left(\text{v}/\text{v}\right)$ is given below as,

$\%\left(\text{v}/\text{v}\right)=\frac{\text{Solute}\text{volume}}{\text{Solution}\text{volume}}\times 100$

The $\text{1-}$ gallon container contains $\text{4}\text{.0}$ quart. Thus, the $\text{5-}$ gallon container contains $\text{4}\times \text{5}=\text{20}$ quart.

The volume of solute and volume of solution is $\text{1}\text{.0}$ quart and $\text{20}\text{.0}$ quart respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

$\begin{array}{rll}\%\left(\text{v}/\text{v}\right)& =& \frac{1}{20}\times 100\\ & =& 5\%\end{array}$

Thus, the concentration in $\%\left(\text{v}/\text{v}\right)$ of $\text{1}\text{.0}$ quart of acetic acid is put into a $\text{5-}$ gallon container is $5\%\left(\text{v}/\text{v}\right)$.

Conclusion

The concentration in $\%\left(\text{v}/\text{v}\right)$ of $\text{1}\text{.0}$ quart of acetic acid is put into a $\text{5-}$ gallon container is $5\%\left(\text{v}/\text{v}\right)$.

Interpretation Introduction

(d)

Interpretation:

The concentration in $\%\left(\text{v}/\text{v}\right)$ of a $\text{300}\text{.-mL}$ sample giving $109\text{mL}$ of acetone on distillation is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume $\left(\text{\%}\text{w}/\text{v}\right)$, percent volume by volume $\left(\text{\%}\text{v}/\text{v}\right)$ and molarity $\left(\text{M}\right)$. When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight $\left(\text{\%}\text{w}/\text{w}\right)$.

Answer to Problem 7.35E

The concentration in $\%\left(\text{v}/\text{v}\right)$ of a $\text{300}\text{.-mL}$ sample giving $109\text{mL}$ of acetone on distillation is $36.33\%\text{v}/\text{v}$.

Explanation of Solution

The formula to calculate concentration in $\%\left(\text{v}/\text{v}\right)$ is given below as,

$\%\left(\text{v}/\text{v}\right)=\frac{\text{Solute}\text{volume}}{\text{Solution}\text{volume}}\times 100$

The volume of acetone and volume of solution is $109\text{mL}$ and $\text{300}\text{.-mL}$ respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

$\begin{array}{rll}\%\left(\text{v}/\text{v}\right)& =& \frac{109\text{mL}}{300\text{mL}}\times 100\\ & =& 36.33\%\text{v}/\text{v}\end{array}$

Thus, the concentration in $\%\left(\text{v}/\text{v}\right)$ of a $\text{300}\text{.-mL}$ sample giving $109\text{mL}$ of acetone on distillation is $36.33\%\text{v}/\text{v}$.

Conclusion

The concentration in $\%\left(\text{v}/\text{v}\right)$ of a $\text{300}\text{.-mL}$ sample giving $109\text{mL}$ of acetone on distillation is $36.33\%\text{v}/\text{v}$.