Chapter 14, Problem 34E

Interpretation Introduction

(a)

Interpretation:

The mass/mass percent concentration of the basic solution $3.00M\text{ NaOH}$ with density, $d=1.12\text{g}/\text{mL}$ is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute per unit volume in liters of solution.

The formula of molarity is given below.

$\text{Molarity}=\frac{\text{Number of moles }\left(n\right)}{\text{Volume in liters (}V\text{)}}$

The density is defined as the mass per unit volume of the compound and is given by the formula given below.

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

Answer to Problem 34E

The mass/mass percent of $3.00M\text{ NaOH}$ solution is $10.71\%$.

Explanation of Solution

The molarity of $\text{ NaOH}$ solution is given $3.00M$.

The density of $\text{ NaOH}$ solution is given $1.12\text{g}/\text{mL}$.

The formula of density is shown below.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ …(1)

The formula of molarity is shown below.

$\text{Molarity}=\frac{\text{Number of moles }\left(n\right)}{\text{Volume in liters (}V\text{)}}$ …(2)

The mass/mass percent is defined as the mass of solute divided by the total mass of the solution multiplied by $100$.

The formula of mass/mass percent is given below.

$\text{Mass}/\text{mass}\text{percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\%$ …(3)

The $3.00M$ solution means three moles of solute in one liter of solution.

Therefore, there are three moles of $\text{ NaOH}$ in one liter of solution.

Mass of solution can be found using the formula of density. Mass is equal to the density multiplied by volume.

Substitue the volume of solution and density in equation (1) to find out the mass of solution.

$\begin{array}{l}\text{Density}=\frac{\text{Mass}}{\text{Volume}}\\ 1.12\text{g}/\text{mL}=\frac{{\rm M}\text{ass}}{1\text{ L}}\end{array}$

Rearrange above equation to find out mass as shown below.

$\begin{array}{rll}{\rm M}\text{ass}& =& 1.12\text{g}/\text{mL}\times 1\text{ L}\\ & =& \text{1}\text{.12 }\text{g}/\text{mL}\times 1000\text{ mL}(\because 1\text{ L}& =& \text{1000 mL})\\ & =& 1120\text{ g}\end{array}$

The mass of solute from its moles can be obtained by multiplying its number of moles by the molar mass of the solute.

The molar mass of $\text{ NaOH}$ is $40.0\text{g}/\text{mol}$.

Mass of three moles of $\text{ NaOH}$ is then $3\text{ mol}\times 40.0\text{g}/\text{mol}=120\text{ g}$.

Substitute the mass of solute and mass of solution in the equation (3) to find out mass/mass percent as shown below.

$\begin{array}{rll}\text{Mass}/\text{mass}\text{percent}& =& \frac{120\text{ g}}{1120\text{ g}}\times 100\%\\ & =& 10.71\%\end{array}$

The mass/mass percent concentration of $3.00M\text{ NaOH}$ solution is $10.71\%$.

Conclusion

The mass/mass percent of $3.00M\text{ NaOH}$ solution is calculated as $10.71\%$.

Interpretation Introduction

(b)

Interpretation:

The mass/mass percent concentration of the basic solution $0.500M\text{ KOH}$ with density, $d=1.02\text{g}/\text{mL}$ is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute per unit volume in liters of solution.

The formula of molarity is given below.

$\text{Molarity}=\frac{\text{Number of moles }\left(n\right)}{\text{Volume in liters (}V\text{)}}$

The density is defined as the mass per unit volume of the compound and is given by the formula given below.

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

Answer to Problem 34E

The mass/mass percent of $0.500M\text{ KOH}$ solution is $2.75\%$.

Explanation of Solution

The molarity of $\text{KOH}$ solution is given $0.500M$.

The density of $\text{KOH}$ solution is given $1.02\text{g}/\text{mL}$.

The formula of density is shown below.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ …(1)

The formula of molarity is shown below.

$\text{Molarity}=\frac{\text{Number of moles }\left(n\right)}{\text{Volume in liters (}V\text{)}}$ …(2)

The mass/mass percent is defined as the mass of solute divided by the total mass of the solution multiplied by $100$.

The formula of mass/mass percent is given below.

$\text{Mass}/\text{mass}\text{percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\%$ …(3)

The $0.500M$ solution means half moles of solute in one liter of solution.

Therefore, there are half moles of $\text{KOH}$ present in one liter of solution.

Mass of solution can be found using the formula of density. Mass is equal to the density multiplied by volume.

Substituent the volume of solution and density in equation (1) to find out the mass of the solution.

$\begin{array}{l}\text{Density}=\frac{\text{Mass}}{\text{Volume}}\\ 1.02\text{g}/\text{mL}=\frac{{\rm M}\text{ass}}{1\text{ L}}\end{array}$

Rearrange above equation to find out mass as shown below.

$\begin{array}{rll}{\rm M}\text{ass}& =& 1.02\text{g}/\text{mL}\times 1\text{ L}\\ & =& \text{1}\text{.02 }\text{g}/\text{mL}\times 1000\text{ mL}(\because 1\text{ L}& =& \text{1000 mL})\\ & =& 1020\text{ g}\end{array}$

The mass of solute from its moles can be obtained by multiplying its number of moles by the molar mass of the solute.

The molar mass of $\text{KOH}$ is $56.11\text{g}/\text{mol}$.

Mass of $0.500$ moles of $\text{KOH}$ is then $0.500\text{ mol}\times 56.11\text{g}/\text{mol}=28.06\text{ g}$.

Substitute the mass of solute and mass of solution in the equation (3) to find out mass/mass percent as shown below.

$\begin{array}{rll}\text{Mass}/\text{mass}\text{percent}& =& \frac{28.06\text{ g}}{1020\text{ g}}\times 100\%\\ & =& 2.75\%\end{array}$

The mass/mass percent of $0.500M\text{ KOH}$ solution is $2.75\%$.

Conclusion

The mass/mass percent of $0.500M\text{ KOH}$ solution is calculated as $2.75\%$.

Interpretation Introduction

(c)

Interpretation:

The mass/mass percent concentration of the basic solution $6.00M{\text{ NH}}_3$ with density, $d=0.954\text{g}/\text{mL}$ is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute per unit volume in liters of solution.

The formula of molarity is given below.

$\text{Molarity}=\frac{\text{Number of moles }\left(n\right)}{\text{Volume in liters (}V\text{)}}$

The density is defined as the mass per unit volume of the compound and is given by the formula given below.

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

Answer to Problem 34E

The mass/mass percent of $6.00M{\text{ NH}}_3$ solution is $10.72\%$.

Explanation of Solution

The molarity of ${\text{NH}}_3$ solution is given $6.00M$.

The density of ${\text{NH}}_3$ solution is given $0.954\text{g}/\text{mL}$.

The formula of density is shown below.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ …(1)

The formula of molarity is shown below.

$\text{Molarity}=\frac{\text{Number of moles }\left(n\right)}{\text{Volume in liters (}V\text{)}}$ …(2)

The mass/mass percent is defined as the mass of solute divided by the total mass of the solution multiplied by $100$.

The formula of mass/mass percent is given below.

$\text{Mass}/\text{mass}\text{percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\%$ …(3)

The $\text{6}\text{.00 }M$ solution means six moles of solute in one liter of solution.

Therefore, there are six moles of ${\text{NH}}_3$ in one liter of solution.

Mass of solution can be found using the formula of density. Mass is equal to the density multiplied by volume.

Substituent the volume of solution and density in equation (1) to find out the mass of the solution.

$\begin{array}{l}\text{Density}=\frac{\text{Mass}}{\text{Volume}}\\ 0.954\text{g}/\text{mL}=\frac{{\rm M}\text{ass}}{1\text{ L}}\end{array}$

Rearrange above equation to find out mass as shown below.

$\begin{array}{rll}{\rm M}\text{ass}& =& 0.954\text{g}/\text{mL}\times 1\text{ L}\\ & =& 0.954\text{g}/\text{mL}\times 1000\text{ mL}(\because 1\text{ L}& =& \text{1000 mL})\\ & =& 954\text{ g}\end{array}$

The mass of solute from its moles can be obtained by multiplying its number of moles by the molar mass of the solute.

The molar mass of ${\text{NH}}_3$ is $17.04\text{g}/\text{mol}$.

Mass of three moles of ${\text{NH}}_3$ is then $6\text{ mol}\times 17.04\text{g}/\text{mol}=102.24\text{ g}$.

Substitute the mass of solute and mass of solution in the equation (3) to find out mass/mass percent as shown below.

$\begin{array}{rll}\text{Mass}/\text{mass}\text{percent}& =& \frac{102.24\text{ g}}{\text{954 g}}\times 100\%\\ & =& 10.72\%\end{array}$

The mass/mass percent of $6.00M{\text{ NH}}_3$ solution is $10.72\%$.

Conclusion

The mass/mass percent of $6.00M{\text{ NH}}_3$ solution is calculated as $10.72\%$.

Interpretation Introduction

(d)

Interpretation:

The mass/mass percent concentration of the basic solution $1.00M{\text{ Na}}_2{\text{CO}}_3$ with density, $d=1.10\text{g}/\text{mL}$ is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute per unit volume in liters of solution.

The formula of molarity is given below.

$\text{Molarity}=\frac{\text{Number of moles }\left(n\right)}{\text{Volume in liters (}V\text{)}}$

The density is defined as the mass per unit volume of the compound and is given by the formula given below.

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

Answer to Problem 34E

The mass/mass percent of $1.00M{\text{ Na}}_2{\text{CO}}_3$ solution is $9.64\%$.

Explanation of Solution

The molarity of ${\text{Na}}_2{\text{CO}}_3$ solution is given $1.00M$.

The density of ${\text{Na}}_2{\text{CO}}_3$ solution is given $1.10\text{g}/\text{mL}$.

The formula of density is shown below.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ …(1)

The formula of molarity is shown below.

$\text{Molarity}=\frac{\text{Number of moles }\left(n\right)}{\text{Volume in liters (}V\text{)}}$ …(2)

The mass/mass percent is defined as the mass of solute divided by the total mass of the solution multiplied by $100$.

The formula of mass/mass percent is given below.

$\text{Mass}/\text{mass}\text{percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\%$ …(3)

The $\text{1}\text{.00 }M$ solution means one mole of solute in one liter of solution.

Therefore, there are one mole of ${\text{NH}}_3$ in one liter of solution.

Mass of solution can be found using the formula of density. Mass is equal to density multiplied by volume.

Substituent the volume of solution and density in equation (1) to find out the mass of the solution.

$\begin{array}{l}\text{Density}=\frac{\text{Mass}}{\text{Volume}}\\ 1.10\text{g}/\text{mL}=\frac{{\rm M}\text{ass}}{1\text{ L}}\end{array}$

Rearrange above equation to find out mass as shown below.

$\begin{array}{rll}{\rm M}\text{ass}& =& 1.10\text{g}/\text{mL}\times 1\text{ L}\\ & =& 1.10\text{g}/\text{mL}\times 1000\text{ mL}(\because 1\text{ L}& =& \text{1000 mL})\\ & =& 1100\text{ g}\end{array}$

The mass of solute from its moles can be obtained by multiplying its number of moles by the molar mass of the solute.

The molar mass of ${\text{Na}}_2{\text{CO}}_3$ is $105.99\text{g}/\text{mol}$.

Mass of three moles of ${\text{Na}}_2{\text{CO}}_3$ is then $1\text{ mol}\times 105.99\text{g}/\text{mol}=105.99\text{ g}$.

Substitute the mass of solute and mass of solution in the equation (3) to find out mass/mass percent as shown below.

$\begin{array}{rll}\text{Mass}/\text{mass}\text{percent}& =& \frac{\text{105}\text{.99 g}}{\text{1100 g}}\times 100\%\\ & =& 9.64\%\end{array}$

The mass/mass percent of $1.00M{\text{ Na}}_2{\text{CO}}_3$ solution is $9.64\%$.

Conclusion

The mass/mass percent of $1.00M{\text{ Na}}_2{\text{CO}}_3$ solution is calculated as $9.64\%$.