Chapter 15, Problem 19E

Interpretation Introduction

(a)

Interpretation:

A concept map is to be drawn and the grams of ${\text{CO}}_2$ gas dissolved in the solution when $\text{1}\text{.0}\text{L}$ of carbon dioxide dissolves in $\text{500}\text{.0 mL}$ of solution is to be calculated.

Concept introduction:

A mole is a basic unit used in the International system of units (SI). It is abbreviated as $\text{mol}$ Mole is defined that the amount of substance that contains molecules or atoms equals to $\text{12}\text{g}$ of $\text{C}-\text{12}$ molecule. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry.

Answer to Problem 19E

The concept map is shown below.

The grams of ${\text{CO}}_2$ gas dissolved in the solution when $\text{1}\text{.0}\text{L}$ of carbon dioxide dissolves in $\text{500}\text{.0 mL}$ of solution is $\text{1}\text{.96}\text{g}$.

Explanation of Solution

When $\text{1}\text{.0}\text{L}$ of carbon dioxide dissolves in $\text{500}\text{.0 mL}$ of solution, mole concept map looks like as shown below.

Figure 1

The volume occupied by $\text{1}\text{mol}$ of ${\text{CO}}_2$ gas at STP is $\text{22}\text{.4}\text{L}$.

Therefore, the number of moles which occupy $\text{1}\text{.0}\text{L}$ is calculated below.

$\begin{array}{rll}\text{Number}\text{of}\text{moles}& =& \text{1}\text{.00}\text{L}{\text{CO}}_{\text{2}}\times \frac{\text{1}\text{mol}{\text{CO}}_{\text{2}}}{\text{22}\text{.4}\text{L}{\text{CO}}_{\text{2}}}\\ & =& \text{0}\text{.04464}\text{mol}\end{array}$

The molar mass of ${\text{CO}}_2$ gas is $44.01\text{g}{\text{mol}}^{-1}$.

Therefore, the mass of $\text{1}\text{mol}$ of ${\text{CO}}_2$ gas is $44.01\text{g}$.

The formula to calculate the mass of $\text{0}\text{.04464}\text{mol}$ of ${\text{CO}}_2$ gas is shown below.

$\text{Mass}= \text{0}\text{.04464}\text{mol}{\text{CO}}_2\times \frac{\text{Mass}\text{of}\text{1}\text{mol}{\text{CO}}_2}{\text{1}\text{mol}{\text{CO}}_2}$

Substitute the mass of $\text{1}\text{mol}$ of ${\text{CO}}_2$ gas in the above equation.

$\begin{array}{rll}\text{Mass}& =& \text{0}\text{.04464}\text{mol}{\text{CO}}_2\times \frac{44.01\text{g}{\text{CO}}_2}{\text{1}\text{mol}{\text{CO}}_2}\\ & =& \text{1}\text{.96}\text{g}\end{array}$

Therefore, the grams of ${\text{CO}}_2$ gas dissolved in the solution is $\text{1}\text{.96}\text{g}$.

Conclusion

The grams of ${\text{CO}}_2$ gas is $\text{1}\text{.96}\text{g}$.

Interpretation Introduction

(b)

Interpretation:

A concept map is to be drawn and the molecules of ${\text{CO}}_2$ gas dissolved in the solution when $\text{1}\text{.0}\text{L}$ of carbon dioxide dissolves in $\text{500}\text{.0 mL}$ of solution is to be calculated.

Concept introduction:

A mole is a basic unit used in the International system of units (SI). It is abbreviated as $\text{mol}$ Mole is defined that the amount of substance that contains molecules or atoms equals to $\text{12}\text{g}$ of $\text{C}-\text{12}$ molecule. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry.

Answer to Problem 19E

The concept map is shown below.

The molecules of ${\text{CO}}_2$ gas dissolved in the solution when $\text{1}\text{.0}\text{L}$ of carbon dioxide dissolves in $\text{500}\text{.0 mL}$ of solution is $\text{2}\text{.69}\times {10}^{22}\text{moleules}$.

Explanation of Solution

When $\text{1}\text{.0}\text{L}$ of carbon dioxide dissolves in $\text{500}\text{.0 mL}$ of solution, mole concept map looks like as shown below.

Figure 1

The volume occupied by $\text{1}\text{mol}$ of ${\text{CO}}_2$ gas at STP is $\text{22}\text{.4}\text{L}$.

Therefore, the number of moles which occupy $\text{1}\text{.0}\text{L}$ of is calculated below.

$\begin{array}{rll}\text{Number}\text{of}\text{moles}& =& \text{1}\text{.00}\text{L}{\text{CO}}_{\text{2}}\times \frac{\text{1}\text{mol}{\text{CO}}_{\text{2}}}{\text{22}\text{.4}\text{L}{\text{CO}}_{\text{2}}}\\ & =& \text{0}\text{.04464}\text{mol}\end{array}$

The molecules present in $\text{1}\text{mol}$ of ${\text{CO}}_2$ gas at STP are $\text{6}\text{.02}\times {10}^{23}\text{moleules}$.

The formula to calculate the molecules occupied by $\text{0}\text{.04464}\text{mol}$ of ${\text{CO}}_2$ gas is shown below.

$\text{Number}\text{of}\text{molecules}= \text{0}\text{.04464}\text{mol}{\text{CO}}_2\times \frac{\text{Molecules}\text{in}\text{1}\text{mol}{\text{CO}}_2}{\text{1}\text{mol}{\text{CO}}_2}$

Substitute the molecules in $\text{1}\text{mol}$ of ${\text{CO}}_2$ gas in the above equation.

$\begin{array}{rll}\text{Number}\text{of}\text{molecules}& =& \text{0}\text{.04464}\text{mol}{\text{CO}}_2\times \frac{\text{6}\text{.02}\times {10}^{23}\text{moleules}}{\text{1}\text{mol}{\text{CO}}_2}\\ & =& \text{2}\text{.69}\times {10}^{22}\text{moleules}\end{array}$

Therefore, the molecules of ${\text{CO}}_2$ gas dissolved in the solution is $\text{2}\text{.69}\times {10}^{22}\text{moleules}$.

Conclusion

The molecules of ${\text{CO}}_2$ gas is $\text{2}\text{.69}\times {10}^{22}\text{moleules}$.

Interpretation Introduction

(c)

Interpretation:

A concept map is to be drawn and the molar concentration of the carbonic acid solution when $\text{1}\text{.0}\text{L}$ of carbon dioxide dissolves in $\text{500}\text{.0 mL}$ of solution is to be calculated.

Concept introduction:

A mole is a basic unit used in the International system of units (SI). It is abbreviated as $\text{mol}$ Mole is defined that the amount of substance that contains molecules or atoms equals to $\text{12}\text{g}$ of $\text{C}-\text{12}$ molecule. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry.

Answer to Problem 19E

The concept map is shown below.

The molar concentration of the carbonic acid solution is $0.0893M$.

Explanation of Solution

When $\text{1}\text{.0}\text{L}$ of carbon dioxide dissolves in $\text{500}\text{.0 mL}$ of solution, mole concept map looks like as shown below.

Figure 1

The volume occupied by $\text{1}\text{mol}$ of ${\text{CO}}_2$ gas at STP is $\text{22}\text{.4}\text{L}$.

Therefore, the number of moles which occupy $\text{1}\text{.0}\text{L}$ of is calculated below.

$\begin{array}{rll}\text{Number}\text{of}\text{moles}& =& \text{1}\text{.00}\text{L}{\text{CO}}_{\text{2}}\times \frac{\text{1}\text{mol}{\text{CO}}_{\text{2}}}{\text{22}\text{.4}\text{L}{\text{CO}}_{\text{2}}}\\ & =& \text{0}\text{.04464}\text{mol}\end{array}$

The number of moles in $\text{500}\text{.0 mL}$ of solution is $\text{0}\text{.04464}\text{mol}$.

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

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

The probable unit factors are given below.

$\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\text{and}\frac{\text{1000}\text{mL}}{\text{1}\text{L}}$

The unit factor to determine $\text{L}$ from $\text{mL}$ is given below.

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

Therefore, the volume in $\text{L}$ is calculated below.

$\begin{array}{rll}\text{Volume}& =& \text{500}\text{.0 mL}\times \frac{\text{1}\text{L}}{\text{1000}\text{mL}}\\ & =& 0.5\text{L}\end{array}$

The formula to determine molarity is shown below.

$M=\frac{\text{n}}{\text{V}}$ …(1)

Where

• $M$ is the molarity of a solution.

• $\text{n}$ is the number of moles of solute.

• $\text{V}$ is the volume of the solution.

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

$\begin{array}{rll}\text{Molar}\text{concentration}\text{of}{\text{CO}}_{\text{2}}& =& \frac{\text{0}\text{.04464}\text{mol}}{\text{0}\text{.5 L}}\\ & =& 0.0893\text{mol}/\text{L}\end{array}$

The relation between $M$ and $\text{mol}/\text{L}$ as shown below.

$1M=1\text{mol}/\text{L}$

The unit factors are given below.

$\frac{1M}{1\text{mol}/\text{L}}\text{and}\frac{1\text{mol}/\text{L}}{1M}$

The unit factor to determine $M$ from $\text{mol}/\text{L}$ is given below.

$\frac{1M}{1\text{mol}/\text{L}}$

Therefore, $0.0893\text{mol}/\text{L}$ can be written as shown below.

$\begin{array}{rll}\text{Molarity}& =& 0.0893\text{mol}/\text{L}\times \frac{1M}{1\text{mol}/\text{L}}\\ & =& 0.0893M\end{array}$

Therefore, the molar concentration of ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$, carbonic acid solution is $0.0893M$.

Conclusion

The molar concentration of carbonic acid solution is $0.0893M$.