Chapter 7, Problem 7.42E

Interpretation Introduction

(a)

Interpretation:

The preparation of the given solutions using pure solute and wateris 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 7.42E

$4.26\text{g}$ of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ is mixed with $\text{200}\text{mL}$ of water to obtain the final solution.

Explanation of Solution

The number of moles of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ 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 $\text{200}\text{mL}$ and $\text{0}\text{.150}\text{M}$ respectively.

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

$\begin{array}{rll}\text{Number}\text{of}\text{moles}& =& \frac{0.150\text{M}\times 200\text{mL}\times 1\text{L}}{1000\text{mL}}\\ \text{Number}\text{of}\text{moles}& =& 0.03\text{mol}\end{array}$

Thus, the number of moles of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ is $0.03\text{mol}$.

The amount of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ is calculated by the formula,

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

The molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ is $142.04\text{g}/\text{mol}$.

Substitute the value of molar mass in the given formula.

$\begin{array}{rll}0.03\text{mol}& =& \frac{\text{Amount}\text{of}{\text{Na}}_{\text{2}}{\text{SO}}_4}{142.04\text{g}/\text{mol}}\\ \text{Amount}\text{of}{\text{Na}}_{\text{2}}{\text{SO}}_4& =& 0.03\text{mol}\times 142.04\text{g}/\text{mol}\\ & =& 4.26\text{g}\end{array}$

Thus, the amount of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ is $4.26\text{g}$.

Hence, $4.26\text{g}$ of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ is mixed with $\text{200}\text{mL}$ of water to obtain the final solution.

Conclusion

$4.26\text{g}$ of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ is mixed with $\text{200}\text{mL}$ 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 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 7.42E

$11.8\text{g}$ of $\text{Zn}{\left({\text{NO}}_3\right)}_2$ is mixed with $\text{250}\text{mL}$ of water to obtain the final solution.

Explanation of Solution

The number of moles of $\text{Zn}{\left({\text{NO}}_3\right)}_2$ 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 $\text{0}\text{.250}\text{M}$ respectively.

Substitute the volume and molarity in equation (1).

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{0.250\text{M}\times 250\text{mL}\times 1\text{L}}{1000\text{mL}}\\ \text{Number}\text{of}\text{moles}=0.0625\text{mol}\end{array}$

Thus, the number of moles of $\text{Zn}{\left({\text{NO}}_3\right)}_2$ is $0.0625\text{mol}$.

The amount of $\text{Zn}{\left({\text{NO}}_3\right)}_2$ is calculated by the formula,

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

The molar mass of $\text{Zn}{\left({\text{NO}}_3\right)}_2$ is $189.4\text{g}/\text{mol}$.

Substitute the value of molar mass in the given formula.

$\begin{array}{rll}0.0625\text{mol}& =& \frac{\text{Amount}\text{of}\text{Zn}{\left({\text{NO}}_3\right)}_2}{189.4\text{g}/\text{mol}}\\ \text{Amount}\text{of}\text{Zn}{\left({\text{NO}}_3\right)}_2& =& 0.0625\text{mol}\times 189.4\text{g}/\text{mol}\\ & =& 11.8\text{g}\end{array}$

Thus, the amount of $\text{Zn}{\left({\text{NO}}_3\right)}_2$ is $11.8\text{g}$.

Hence, $11.8\text{g}$ of $\text{Zn}{\left({\text{NO}}_3\right)}_2$ is mixed with $\text{250}\text{mL}$ of water to obtain the final solution.

Conclusion

$11.8\text{g}$ of $\text{Zn}{\left({\text{NO}}_3\right)}_2$ is mixed with $\text{250}\text{mL}$ 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.

Concept Introduction:

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

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

Answer to Problem 7.42E

$3.4\text{g}$ of sodium chloride is mixed with $\text{146}\text{.6}\text{mL}$ 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,

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

The given value of $\%\left(\text{w}/\text{w}\right)$ is $2.25\%\left(\text{w}/\text{w}\right)$. The mass of solution is $\text{150}\text{g}$.

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

$\begin{array}{rll}2.25& =& \frac{\text{solute}\text{mass}}{150\text{g}}\times 100\\ \text{solute}\text{mass}& =& \frac{150\times 2.25}{100}\\ & =& 3.4\text{g}\end{array}$

Hence, the mass of solute is $3.4\text{g}$.

Mass of solvent is calculated as follows:

$\text{Mass}\text{of}\text{solution}=\text{Mass}\text{of}\text{solute}+\text{Mass}\text{of}\text{solvent}$

Substitute the mass of solute and solution in above formula.

$\begin{array}{rll}\text{150}\text{g}& =& \text{3}\text{.4}\text{g}+\text{Mass}\text{of}\text{solvent}\\ \text{Mass}\text{of}\text{solvent}& =& \left(\text{150}-\text{3}\text{.4}\right)\text{g}\\ & =& 146.6\text{g}\end{array}$

Thus, the mass of solvent (water) is $146.6\text{g}$. The density of water is $\text{1}\text{.00}\text{g}/\text{mL}$. Hence, the volume of water is $\text{146}\text{.6}\text{mL}$.

Therefore, $3.4\text{g}$ of sodium chloride is mixed with $\text{146}\text{.6}\text{mL}$ of water to obtain the final solution.

Conclusion

$3.4\text{g}$ of sodium chloride is mixed with $\text{146}\text{.6}\text{mL}$ of water 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 $\%\left(\text{w}/\text{v}\right)$ 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 7.42E

$0.94\text{g}$ of $\text{KCl}$ is mixed with $\text{125}\text{mL}$ of water 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 $0.75\%\left(\text{w}/\text{v}\right)$. The volume of $\text{KCl}$ solution is $\text{125}\text{mL}$.

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

$\begin{array}{rll}0.75& =& \frac{\text{grams}\text{of}\text{solute}}{\text{125}\text{mL}}\times 100\\ \text{grams}\text{of}\text{solute}& =& \frac{0.75\times 125\text{mL}}{100}\\ & =& 0.94\text{g}\end{array}$

Hence, $0.94\text{g}$ of $\text{KCl}$ is mixed with $\text{125}\text{mL}$ of water to obtain the final solution.

Conclusion

$0.94\text{g}$ of $\text{KCl}$ is mixed with $\text{125}\text{mL}$ of water to obtain the final solution.