Chapter 7, Problem 7.43E

Explain how you would prepare the following solutions using pure solute and water. Assume water has a density of $\text{1}\text{.00}\text{g}/\text{mL}$.

a. $\text{250}\text{mL}$ of a $\text{2}\text{.00}\text{M}$ $\text{NaOH}$ solution.

b. $\text{500}\text{mL}$ of a $40.0\%\left(\text{v}/\text{v}\right)$ alcohol solution $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)$

c. $\text{100}\text{mL}$ of $15.0\%\left(\text{w}/\text{v}\right)$ glycerol solution. Glycerol is a liquid with a density of $\text{1}\text{.26}\text{g}/\text{mL}$. Describe two ways to measure out the amount of glycerol needed.

d. Approximately $\text{50}\text{mL}$ of a normal saline solution, $\text{0}\text{.89\%}\left(\text{w}/\text{w}\right)\text{NaCl}$.

Interpretation Introduction

(a)

Interpretation:

The preparation of the given solutions using pure solute and water is to be explained.

Concept Introduction:

The number of moles is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$

Answer to Problem 7.43E

$2.00\text{g}$ of $\text{NaOH}$ is mixed with $\text{250}\text{mL}$ of water to obtain the final solution.

Explanation of Solution

The number of moles of $\text{NaOH}$ is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\left(\text{mL}\right)}\times \frac{1000\text{mL}}{1\text{L}}$

The above formula can be written as follows:

$\text{Number}\text{of}\text{moles}=\frac{\text{Molarity}\times \text{Volume}\left(\text{mL}\right)\times 1\text{L}}{1000\text{mL}}$ … (1)

The given volume and molarity is $\text{250}\text{mL}$ and $\text{2}\text{.00}\text{M}$ respectively.

Substitute the volume and molarity in the equation (1).

$\begin{array}{rll}\text{Number}\text{of}\text{moles}& =& \frac{0.200\text{M}\times 250\text{mL}\times 1\text{L}}{1000\text{mL}}\\ \text{Number}\text{of}\text{moles}& =& 0.05\text{mol}\end{array}$

Thus, the number of moles of $\text{NaOH}$ is $0.05\text{mol}$.

The amount of $\text{NaOH}$ is calculated by the formula,

$\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

The molar mass of $\text{NaOH}$ is $40\text{g}/\text{mol}$.

Substitute the value of molar mass in the given formula.

$\begin{array}{rll}0.05\text{mol}& =& \frac{\text{Amount}\text{of}\text{NaOH}}{40\text{g}/\text{mol}}\\ \text{Amount}\text{of}\text{NaOH}& =& 0.05\text{mol}\times 40\text{g}/\text{mol}\\ & =& 2.00\text{g}\end{array}$

Thus, the amount of $\text{NaOH}$ is $2.00\text{g}$.

Hence, $2.00\text{g}$ of $\text{NaOH}$ is mixed with $\text{250}\text{mL}$ of water to obtain the final solution.

Conclusion

$2.00\text{g}$ of $\text{NaOH}$ is mixed with $\text{250}\text{mL}$ of water to obtain the final solution.

Interpretation Introduction

(b)

Interpretation:

The preparation of the given solutions using pure solute and water is to be explained.

Concept Introduction:

The concentration of the solution in $\text{\%}\left(\text{v}/\text{v}\right)$ is given by the formula,

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

Answer to Problem 7.43E

$200\text{mL}$ of alcohol solution is mixed with $\text{500}\text{mL}$ of water to obtain the final solution.

Explanation of Solution

The concentration of the solution in $\%\left(\text{v}/\text{v}\right)$ is given by the formula,

$\text{\%}\left(\text{v}/\text{v}\right)=\frac{\text{solute}\text{volume}}{\text{solution}\text{volume}}\times 100$ … (1)

The given value of $\%\left(\text{v}/\text{v}\right)$ is $40.0\%\left(\text{v}/\text{v}\right)$. The volume of solution is $\text{500}\text{mL}$.

Substitute the value of $\%\left(\text{v}/\text{v}\right)$ and volume of solution in equation (1).

$\begin{array}{rll}40.0& =& \frac{\text{solute}\text{volume}}{500\text{mL}}\times 100\\ \text{solute}\text{volume }& =& \frac{500\times 40.0}{100}\\ & =& 200\text{mL}\end{array}$

Hence, the volume of solute is $200\text{mL}$.

Therefore, $200\text{mL}$ of alcohol solution is mixed with $\text{500}\text{mL}$ of water to obtain the final solution.

Conclusion

$200\text{mL}$ of alcohol solution is mixed with $\text{500}\text{mL}$ of water to obtain the final solution.

Interpretation Introduction

(c)

Interpretation:

The preparation of the given solutions using pure solute and water is to be explained. The two ways to measure out the amount of glycerol is to be described.

Concept Introduction:

The concentration of the solution in $\%\left(\text{w}/\text{v}\right)$ is given by the formula,

$\%\left(\text{w}/\text{v}\right)=\frac{\text{grams}\text{of}\text{solute}}{\text{milliliters}\text{of}\text{solution}}\times 100$

Answer to Problem 7.43E

$15.0\text{g}$ of glycerol is mixed with $\text{100}\text{mL}$ of water to obtain the final solution. $11.90\text{mL}$ of glycerol along with remaining amount of water, is added to obtain the final solution.

Explanation of Solution

The concentration of the solution in $\%\left(\text{w}/\text{v}\right)$ is given by the formula,

$\%\left(\text{w}/\text{v}\right)=\frac{\text{grams}\text{of}\text{solute}}{\text{milliliters}\text{of}\text{solution}}\times 100$ …(1)

The given value of $\%\left(\text{w}/\text{v}\right)$ is $15.0\%\left(\text{w}/\text{v}\right)$. The volume of glycerol solution is $\text{100}\text{mL}$.

Substitute the value of $\%\left(\text{w}/\text{v}\right)$ and volume in equation (1).

$\begin{array}{rll}15.0& =& \frac{\text{grams}\text{of}\text{solute}}{\text{100}\text{mL}}\times 100\\ \text{grams}\text{of}\text{solute}& =& \frac{15.0\times 100\text{mL}}{100}\\ & =& 15.0\text{g}\end{array}$

Hence, $15.0\text{g}$ of glycerol is mixed with $\text{100}\text{mL}$ of water to obtain the final solution.

The density of glycerol is $\text{1}\text{.26}\text{g}/\text{mL}$.

The density is given by the formula,

$\text{Density}=\frac{\text{Mass}\text{of}\text{solution}}{\text{Volume}\text{of}\text{solution}}$

Substitute the density and mass of glycerol and in the above formula.

$\begin{array}{rll}\text{1}\text{.26}\text{g}/\text{mL}& =& \frac{\text{15}\text{.0}\text{g}}{\text{volume}\text{of}\text{glycerol}}\\ \text{volume}\text{of}\text{glycerol}& =& \frac{\text{15}\text{.0}\text{g}}{\text{1}\text{.26}\text{g}/\text{mL}}\\ & =& 11.90\text{mL}\end{array}$

Hence, $11.90\text{mL}$ of glycerol along with remaining amount of water, is added to obtain the final solution.

Conclusion

$15.0\text{g}$ of glycerol is mixed with $\text{100}\text{mL}$ of water to obtain the final solution. $11.90\text{mL}$ of glycerol along with remaining amount of water, is added to obtain the final solution.

Interpretation Introduction

(d)

Interpretation:

The preparation of the given solutions using pure solute and water is to be explained.

Concept Introduction:

The concentration of the solution in $\text{\%}\left(\text{w}/\text{w}\right)$ is given by the formula,

$\text{\%}\left(\text{w}/\text{w}\right)=\frac{\text{solute}\text{mass}}{\text{solution}\text{mass}}\times 100$

Answer to Problem 7.43E

$0.445\text{g}$ of sodium chloride is mixed with $\text{50}\text{mL}$ of water to obtain the final solution.

Explanation of Solution

The concentration of the solution in $\text{\%}\left(\text{w}/\text{w}\right)$ is given by the formula,

$\text{\%}\left(\text{w}/\text{w}\right)=\frac{\text{solute}\text{mass}}{\text{solution}\text{mass}}\times 100$ … (1)

The given value of $\text{\%}\left(\text{w}/\text{w}\right)$ is $\text{0}\text{.89\%}\left(\text{w}/\text{w}\right)$. The density of water is $\text{1}\text{g}/\text{mL}$. Thus, the mass of water in $\text{50}\text{mL}$ will be $\text{50}\text{g}$. Therefore, the mass of the solution is $\text{50}\text{g}$.

Substitute the value of $\%\left(\text{w}/\text{w}\right)$ and mass of solution in equation (1).

$\begin{array}{rll}\text{0}\text{.89}& =& \frac{\text{solute}\text{mass}}{\text{50}\text{g}}\times 100\\ \text{solute}\text{mass }& =& \frac{\text{50}\text{g}\times 0.89}{100}\\ & =& 0.445\text{g}\end{array}$

Hence, the mass of solute is $0.445\text{g}$.

Therefore, $0.445\text{g}$ of sodium chloride is mixed with $\text{50}\text{mL}$ of water to obtain the final solution.

Conclusion

$0.445\text{g}$ of sodium chloride is mixed with $\text{50}\text{mL}$ of water to obtain the final solution.