Chapter 15, Problem 22E

Interpretation Introduction

(a)

Interpretation:

The concept map is to be drawn and the volume of sulfur trioxide $\left({\text{SO}}_3\right)$ gas dissolved in $0.325\text{L}$ of solution is to be stated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as present in $12\text{g}$ of $\text{C}-12$. The number of particles present in one mole of a substance is $6.023\times {10}^{23}$ particle. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry. The unit of the mole is denoted as $\text{mol}$.

Answer to Problem 22E

The mole concept map is shown below.

The liters of ${\text{SO}}_3$ gas at $\text{STP}$ dissolved in $\text{0}\text{.325}\text{L}$ of solution is $\text{0}\text{.826}\text{L}$.

Explanation of Solution

It is given that the number of molecules of ${\text{SO}}_3$ is $1.12\times {10}^{23}$ which are dissolved in $\text{0}\text{.325}\text{L}$ of solution.

The mole concept map is shown below.

Figure 1

The Figure 1 shows that the number of moles should be calculated first, to calculate other parameters.

The number of moles is calculated from the relation shown below.

$\begin{array}{rll}1\text{mol}& =& 6.022\times {10}^{23}\text{molecules}\\ \frac{\text{1}}{6.022\times {10}^{23}}\text{mol}& =& 1\text{molecule}\end{array}$

Therefore, the number of moles for $\text{1}{\text{.12×10}}^{\text{23}}\text{molecules}$ is calculated as shown below.

$\begin{array}{rll}\text{The}\text{number}\text{of}\text{moles}\text{of}{\text{SO}}_3& =& \frac{\text{1}\text{mol}}{6.022\times {10}^{23}\text{molecules}}\times \left(1.12\times {10}^{23}\right)\text{molecules}\\ & =& \text{0}\text{.186}\text{mol}\end{array}$

The volume of ${\text{SO}}_3$ gas is calculated from the relation as shown below.

$1\text{mol}=\text{2}\text{.24}\text{L}\text{of}{\text{SO}}_3$

The number of moles of ${\text{SO}}_3$ gas is $\text{0}\text{.186}\text{mol}$.

The volume for $\text{0}\text{.186}\text{mol}$ of ${\text{SO}}_3$ is calculated as shown below

$\begin{array}{rll}\text{Volume}\text{of}{\text{SO}}_3& =& 0.186\text{mol}\times \frac{\text{22}\text{.4}\text{L}}{\text{1}\text{mol}}\\ & =& 4.166\text{L}\end{array}$

The volume of ${\text{SO}}_3$ gas at $\text{STP}$ is $4.166\text{L}$.

Conclusion

The mole concept map is shown Figure 1. The volume of ${\text{SO}}_3$ gas dissolved at $\text{STP}$ is $4.166\text{L}$.

Interpretation Introduction

(b)

Interpretation:

The mole concept map is to be drawn and the mass of ${\text{SO}}_3$ gas dissolved in the solution is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as present in $12\text{g}$ of $\text{C}-12$. The number of particles present in one mole of a substance is $6.023\times {10}^{23}$ particle. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry. The unit of the mole is denoted as $\text{mol}$.

Answer to Problem 22E

The mole concept map is drawn below and the grams of ${\text{SO}}_3$ gas dissolved in $\text{0}\text{.325}\text{L}$ of solution is $14.9\text{g}$.

Explanation of Solution

It is given that the number of molecules of ${\text{SO}}_3$ is $1.12\times {10}^{23}$ which are dissolved in $\text{0}\text{.325}\text{L}$ of solution.

The mole concept map is drawn below.

Figure 1

The above diagram shows that the number of moles should be calculated first, to calculate other parameters.

The number of moles is calculated from the relation shown below.

$\begin{array}{rll}1\text{mol}& =& 6.022\times {10}^{23}\text{molecules}\\ \frac{\text{1}}{6.022\times {10}^{23}}\text{mol}& =& 1\text{molecule}\end{array}$

Therefore, the number of moles for $\text{1}{\text{.12×10}}^{\text{23}}\text{molecules}$ is calculated as shown below.

$\begin{array}{rll}\text{The}\text{number}\text{of}\text{moles}\text{of}{\text{SO}}_3& =& \frac{\text{1}\text{mol}}{6.022\times {10}^{23}\text{molecules}}\times \left(1.12\times {10}^{23}\right)\text{molecules}\\ & =& \text{0}\text{.186}\text{mol}\end{array}$

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-\text{1}}$.

The molar mass of sulfur is $32.07\text{g}{\text{mol}}^{-\text{1}}$.

$\begin{array}{rll}\text{Molar}\text{mass}\text{of}{\text{SO}}_3\text{gas}& =& \text{Mass}\text{of}\text{S}+\left(3\times \text{mass}\text{of}\text{O}\right)\\ & =& 32.07\text{g}{\text{mol}}^{-\text{1}}+\left(3\times 16.00\text{g}{\text{mol}}^{-\text{1}}\right)\\ & =& 32.07\text{g}{\text{mol}}^{-\text{1}}+48.00\text{g}{\text{mol}}^{-\text{1}}\\ & =& 80.07\text{g}{\text{mol}}^{-\text{1}}\end{array}$

The mass of ${\text{SO}}_3$ gas is calculated from the relation as shown below.

$\begin{array}{l}1\text{mol}=\text{Molar}\text{mass}\text{of}{\text{SO}}_3\\ 1\text{mol}=80.07\text{g}\text{of}{\text{SO}}_3\end{array}$

The number of moles of ${\text{SO}}_3$ gas is $\text{0}\text{.186}\text{mol}$.

The mass for $\text{0}\text{.186}\text{mol}$ of ${\text{SO}}_3$ is calculated as shown below.

$\begin{array}{rll}\text{Mass}\text{of}{\text{SO}}_3& =& \text{0}\text{.186}\text{mol×}\frac{80.07\text{g}{\text{SO}}_3}{\text{1}\text{mol}{\text{SO}}_3}\\ & =& 14.89\text{g}{\text{SO}}_3\\ & \approx & 14.9\text{g}{\text{SO}}_3\end{array}$

Therefore, the mass of ${\text{SO}}_3$ gas is $14.9\text{g}$.

Conclusion

The mass of ${\text{SO}}_3$ gas dissolved in $\text{0}\text{.325}\text{L}$ of solution is. $14.9\text{g}$.

Interpretation Introduction

(c)

Interpretation:

The mole concept map is to be drawn and the molar concentration of ${\text{SO}}_3$ solution dissolved in the solution is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as present in $12\text{g}$ of $\text{C}-12$. The number of particles present in one mole of a substance is $6.023\times {10}^{23}$ particle. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry. The unit of the mole is denoted as $\text{mol}$. The unit of the mole is denoted as mol. Molarity is defined as the number of moles in $\text{1000}\text{mL}$ of solution. Molarity is denoted by $M$.

Answer to Problem 22E

The mole concept map is drawn below and the molar concentration of ${\text{SO}}_3$ gas dissolved in $\text{0}\text{.325}\text{L}$ of solution is $0.0819M$.

Explanation of Solution

It is given that the number of molecules of ${\text{SO}}_3$ is $\text{1}{\text{.12×10}}^{\text{23}}$ which are dissolved in $\text{0}\text{.325}\text{L}$ of solution.

The mole concept map is shown below.

Figure 1

The above diagram shows that the number of moles should be calculated first, to calculate other parameters.

The number of moles is calculated from the relation shown below.

$\begin{array}{rll}1\text{mol}& =& 6.022\times {10}^{23}\text{molecules}\\ \frac{\text{1}}{6.022\times {10}^{23}}\text{mol}& =& 1\text{molecule}\end{array}$

Therefore, the number of moles for $\text{1}{\text{.12×10}}^{\text{23}}\text{molecules}$ is calculated as shown below.

$\begin{array}{rll}\text{The}\text{number}\text{of}\text{moles}\text{of}{\text{SO}}_3& =& \frac{\text{1}\text{mol}}{6.022\times {10}^{23}\text{molecules}}\times \left(1.12\times {10}^{23}\right)\text{molecules}\\ & =& \text{0}\text{.186}\text{mol}\end{array}$

The number of moles for of ${\text{SO}}_3$ in $\text{0}\text{.325}\text{L}$ of solution is $\text{0}\text{.0369}\text{mol}$.

The formula to determine molarity is shown below.

$M=\frac{n}{V}$ …(1)

Where

• $M$ is the molarity of a solution.

• $n$ is the number of moles of solute.

• $V$ is the volume of the solution.

Substitute the value of number of moles as $\text{0}\text{.186}\text{mol}$ and volume as $\text{0}\text{.325}\text{L}$ in equation (1).

$\begin{array}{rll}\text{Molarity}\text{of}{\text{SO}}_{\text{3}}& =& \frac{\text{0}\text{.186}\text{mol}}{\text{0}\text{.325L}}\\ & =& \left(0.5723\text{mol}/\text{L}\right)\left(\frac{1M}{1\text{mol}/\text{L}}\right)\\ & =& 0.5723M\end{array}$

Therefore, the molar concentration of ${\text{SO}}_3$ gas is $0.5723M$.

Conclusion

The molar concentration of ${\text{SO}}_3$ gas dissolved in $\text{0}\text{.325}\text{L}$ of solution is. $0.5723M$.