Chapter 14, Problem 33E

Interpretation Introduction

(a)

Interpretation:

The $\text{mass}/\text{mass}$ percent of the $6.00M\text{ HCl}$ is to be calculated.

Concept introduction:

Solution is a homogenous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. Molarity is defined as the number of moles of solute present in one liter of the solution. Mass percent is defined as the mass of the solute in $\text{100}\text{g}$ if the solution.

Answer to Problem 33E

The $\text{mass}/\text{mass}$ percent of $6.00M\text{ HCl}$ solution is $19.9\%$.

Explanation of Solution

The formula of density is shown below.

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

The molarity of $\text{HCl}$ solution is $6.00M$.

This means that $1\text{L}$ of the solution contains six moles.

So, the volume of the solution is $1\text{L}$.

The relation between $\text{L}$ and $\text{mL}$ is shown below.

$\text{1}\text{L}=\text{1000}\text{mL}$

The density of $\text{HCl}$ solution is $1.10\text{g}/\text{mL}$.

Substitute the value of density and volume in equation (1).

$\begin{array}{rll}1.10\text{g}/\text{mL}& =& \frac{\text{Mass}}{1\text{L}}\\ \text{Mass}& =& 1.10\text{g}/\text{mL}\times 1000\text{mL}\\ & =& 1100\text{g}\end{array}$

Therefore, the mass of the solution is $1100\text{g}$.

The formula of $\text{mass}/\text{mass}$ percent is given below.

$\text{Mass}/\text{mass}\text{percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\%$ …(2)

The molar mass of $\text{HCl}$ is $36.6\text{ g}{\text{mol}}^{-1}$.

The mass of $1$ mole of $\text{HCl}$ is $36.6\text{ g}$.

The mass of $6$ moles of $\text{HCl}$ is calculated as shown below.

$\begin{array}{rll}\text{Mass}\text{of}\text{solute}& =& 6\times 36.46\text{g}\\ & =& \text{218}\text{.76}\end{array}$

Substitute the mass of solute and mass of solution in the equation (2).

$\begin{array}{rll}\text{Mass}/\text{mass}\text{percent}& =& \frac{\text{218}\text{.76}}{1100\text{g}}\times 100\%\\ & =& 19.9\%\end{array}$

The $\text{mass}/\text{mass}$ percent of $6.00M\text{ HCl}$ solution is $19.9\%$.

Conclusion

The $\text{mass}/\text{mass}$ percent of $6.00M\text{ HCl}$ solution is $19.9\%$.

Interpretation Introduction

(b)

Interpretation:

The $\text{mass}/\text{mass}$ percent of the $1.00M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ is to be calculated.

Concept introduction:

Solution is a homogenous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. Molarity is defined as the number of moles of solute present in one liter of the solution. Mass percent is defined as the mass of the solute in $\text{100}\text{g}$ if the solution.

Answer to Problem 33E

The $\text{mass}/\text{mass}$ percent of $1.00M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ solution is $5.95\%$.

Explanation of Solution

The formula of density is shown below.

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

The molarity of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ solution is $1.00M$.

This means that $1\text{L}$ of the solution contains one mole.

So, the volume of the solution is $1\text{L}$.

The relation between $\text{L}$ and $\text{mL}$ is shown below.

$\text{1}\text{L}=\text{1000}\text{mL}$

The density of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ solution is $1.01\text{g}/\text{mL}$.

Substitute the value of density and volume in equation (1).

$\begin{array}{rll}1.01\text{g}/\text{mL}& =& \frac{\text{Mass}}{1\text{L}}\\ \text{Mass}& =& 1.01\text{g}/\text{mL}\times 1000\text{mL}\\ & =& 1010\text{g}\end{array}$

Therefore, the mass of the solution is $1010\text{g}$.

The formula of $\text{mass}/\text{mass}$ percent is given below.

$\text{Mass}/\text{mass}\text{percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\%$ …(2)

The molar mass of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ is $60.06\text{g}{\text{mol}}^{-1}$.

The mass of $1$ mole of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ is $60.06\text{g}$.

Substitute the mass of solute and mass of solution in the equation (2).

$\begin{array}{rll}\text{Mass}/\text{mass}\text{percent}& =& \frac{60.06\text{g}}{1010\text{g}}\times 100\%\\ & =& 5.95\%\end{array}$

The $\text{mass}/\text{mass}$ percent of $1.00M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ solution is $5.95\%$.

Conclusion

The $\text{mass}/\text{mass}$ percent of $1.00M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ solution is $5.95\%$.

Interpretation Introduction

(c)

Interpretation:

The $\text{mass}/\text{mass}$ percent of the $0.500M{\text{ HNO}}_3$ is to be calculated.

Concept introduction:

Solution is a homogenous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. Molarity is defined as the number of moles of solute present in one liter of the solution. Mass percent is defined as the mass of the solute in $\text{100}\text{g}$ if the solution.

Answer to Problem 33E

The $\text{mass}/\text{mass}$ percent of $0.500M{\text{ HNO}}_3$ solution is $3.12\%$.

Explanation of Solution

The formula of density is shown below.

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

The molarity of ${\text{HNO}}_3$ solution is $0.500M$.

This means that $1\text{L}$ of the solution contains $0.5$ mole.

So, the volume of the solution is $1\text{L}$.

The relation between $\text{L}$ and $\text{mL}$ is shown below.

$\text{1}\text{L}=\text{1000}\text{mL}$

The density of ${\text{HNO}}_3$ solution is $1.01\text{g}/\text{mL}$.

Substitute the value of density and volume in equation (1).

$\begin{array}{rll}1.01\text{g}/\text{mL}& =& \frac{\text{Mass}}{1\text{L}}\\ \text{Mass}& =& 1.01\text{g}/\text{mL}\times 1000\text{mL}\\ & =& 1010\text{g}\end{array}$

Therefore, the mass of the solution is $1010\text{g}$.

The formula of $\text{mass}/\text{mass}$ percent is given below.

$\text{Mass}/\text{mass}\text{percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\%$ …(2)

The molar mass of ${\text{HNO}}_3$ is $63\text{g}{\text{mol}}^{-1}$.

The mass of $1$ mole of ${\text{HNO}}_3$ is $63\text{g}$.

The mass of $0.5$ moles of ${\text{HNO}}_3$ is calculated as shown below.

$\begin{array}{rll}\text{Mass}\text{of}\text{solute}& =& 0.5\times 63\text{g}\\ & =& \text{31}\text{.5}\text{g}\end{array}$

Substitute the mass of solute and mass of solution in the equation (2).

$\begin{array}{rll}\text{Mass}/\text{mass}\text{percent}& =& \frac{31.5\text{g}}{1010\text{g}}\times 100\%\\ & =& 3.12\%\end{array}$

The $\text{mass}/\text{mass}$ percent of $0.500M{\text{ HNO}}_3$ solution is $3.12\%$.

Conclusion

The $\text{mass}/\text{mass}$ percent of $0.500M{\text{ HNO}}_3$ solution is $3.12\%$.

Interpretation Introduction

(d)

Interpretation:

The $\text{mass}/\text{mass}$ percent of the basic solution $3.00M{\text{ H}}_2{\text{SO}}_4$ is to be calculated.

Concept introduction:

Solution is a homogenous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. Molarity is defined as the number of moles of solute per unit volume in liters of solution. Mass percent is defined as the mass of the solute in $\text{100}\text{g}$ if the solution.

Answer to Problem 33E

The $\text{mass}/\text{mass}$ percent of $3.00M{\text{ H}}_2{\text{SO}}_4$ solution is $24.9\%$.

Explanation of Solution

The formula of density is shown below.

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

The molarity of ${\text{H}}_2{\text{SO}}_4$ solution is $3.00M$.

This means that $1\text{L}$ of the solution contains $3$ moles.

So, the volume of the solution is $1\text{L}$.

The relation between $\text{L}$ and $\text{mL}$ is shown below.

$\text{1}\text{L}=\text{1000}\text{mL}$

The density of ${\text{H}}_2{\text{SO}}_4$ solution is $1.18\text{g}/\text{mL}$.

Substitute the value of density and volume in equation (1).

$\begin{array}{rll}1.18\text{g}/\text{mL}& =& \frac{\text{Mass}}{1\text{L}}\\ \text{Mass}& =& 1.18\text{g}/\text{mL}\times 1000\text{mL}\\ & =& 1180\text{g}\end{array}$

Therefore, the mass of the solution is $1180\text{g}$.

The formula of $\text{mass}/\text{mass}$ percent is given below.

$\text{Mass}/\text{mass}\text{percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\%$ …(2)

The molar mass of ${\text{H}}_2{\text{SO}}_4$ is $98\text{g}{\text{mol}}^{-1}$.

The mass of $1$ mole of ${\text{H}}_2{\text{SO}}_4$ is $98\text{g}$.

The mass of $3$ moles of ${\text{H}}_2{\text{SO}}_4$ is calculated as shown below.

$\begin{array}{rll}\text{Mass}\text{of}\text{solute}& =& 3\times 98\text{g}\\ & =& \text{294}\text{g}\end{array}$

Substitute the mass of solute and mass of solution in the equation (2).

$\begin{array}{rll}\text{Mass}/\text{mass}\text{percent}& =& \frac{\text{294}\text{g}}{1180\text{g}}\times 100\%\\ & =& 24.9\%\end{array}$

The $\text{mass}/\text{mass}$ percent of $3.00M{\text{ H}}_2{\text{SO}}_4$ solution is $24.9\%$.

Conclusion

The $\text{mass}/\text{mass}$ percent of $3.00M{\text{ H}}_2{\text{SO}}_4$ solution is $24.9\%$.