Chapter 15, Problem 16E

Interpretation Introduction

(a)

Interpretation:

A concept map is to be drawn and the grams of ${\text{NH}}_3$ gas dissolved in the solution when $\text{0}\text{.0755}\text{mol}$ of ammonia gas dissolves in $\text{0}\text{.155 L}$ 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 16E

The concept map is shown below.

The grams of ${\text{NH}}_3$ gas dissolved in the solution when $\text{0}\text{.0755}\text{mol}$ of ammonia gas dissolves in $\text{0}\text{.155 L}$ of solution is $\text{1}\text{.29}\text{g}$.

Explanation of Solution

When $\text{0}\text{.0755}\text{mol}$ of ammonia gas dissolves in $\text{0}\text{.155 L}$ of solution, mole concept map looks like as shown below.

Figure 1

The molar mass of ${\text{NH}}_3$ gas is $17.04\text{g}{\text{mol}}^{-1}$.

Therefore, the mass of $\text{1}\text{mol}$ of ${\text{NH}}_3$ gas is $17.04\text{g}$.

The formula to calculate the mass of $\text{0}\text{.0755}\text{mol}$ of ${\text{NH}}_3$ gas is shown below.

$\text{Mass}= \text{0}\text{.0755}\text{mol}{\text{NH}}_3\times \frac{\text{Mass}\text{of}\text{1}\text{mol}{\text{NH}}_3}{\text{1}\text{mol}{\text{NH}}_3}$

Substitute the mass of $\text{1}\text{mol}$ of ${\text{NH}}_3$ gas in the above equation.

$\begin{array}{rll}\text{Mass}& =& \text{0}\text{.0755}\text{mol}{\text{NH}}_3\times \frac{17.04\text{g}}{\text{1}\text{mol}{\text{NH}}_3}\\ & =& \text{1}\text{.29}\text{g}\end{array}$

Therefore, the grams of ${\text{NH}}_3$ gas dissolved in the solution is $\text{1}\text{.29}\text{g}$.

Conclusion

The grams of ${\text{NH}}_3$ gas is $\text{1}\text{.29}\text{g}$.

Interpretation Introduction

(b)

Interpretation:

A concept map is to be drawn and the liters of ${\text{NH}}_3$ gas dissolved in the solution in when $\text{0}\text{.0755}\text{mol}$ of ammonia gas dissolves in $\text{0}\text{.155 L}$ 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 16E

The concept map is shown below.

The liters of ${\text{NH}}_3$ gas dissolved in the solution when $\text{0}\text{.0755}\text{mol}$ of ammonia gas dissolves in $\text{0}\text{.155 L}$ of solution is $\text{1}\text{.69}\text{L}$.

Explanation of Solution

When $\text{0}\text{.0755}\text{mol}$ of ammonia gas dissolves in $\text{0}\text{.155 L}$ of solution, mole concept map looks like as shown below.

Figure 1

The volume occupied by $\text{1}\text{mol}$ of ${\text{NH}}_3$ gas at STP is $\text{22}\text{.4}\text{L}$.

The formula to calculate the volume occupied by $\text{0}\text{.0755}\text{mol}$ of ${\text{NH}}_3$ gas is shown below.

$\text{Volume}= \text{0}\text{.0755}\text{mol}{\text{NH}}_3\times \frac{\text{Volume}\text{of}\text{1}\text{mol}{\text{NH}}_3}{\text{1}\text{mol}{\text{NH}}_3}$

Substitute the volume of $\text{1}\text{mol}$ of ${\text{NH}}_3$ gas in the above equation.

$\begin{array}{rll}\text{Volume}& =& \text{0}\text{.0755}\text{mol}{\text{NH}}_3\times \frac{\text{22}\text{.4}\text{L}}{\text{1}\text{mol}{\text{NH}}_3}\\ & =& \text{1}\text{.69}\text{L}\end{array}$

Therefore, the liters of ${\text{NH}}_3$ gas dissolved in the solution is $\text{1}\text{.69}\text{L}$.

Conclusion

The liters of ${\text{NH}}_3$ gas is $\text{1}\text{.69}\text{L}$.

Interpretation Introduction

(c)

Interpretation:

A concept map is to be drawn and the molecules of ${\text{NH}}_3$ gas dissolved in the solution in when $\text{0}\text{.0755}\text{mol}$ of ammonia gas dissolves in $\text{0}\text{.155 L}$ 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 16E

The concept map is shown below.

The molecules of ${\text{NH}}_3$ gas dissolved in the solution when $\text{0}\text{.0755}\text{mol}$ of ammonia gas dissolves in $\text{0}\text{.155 L}$ of solution is $\text{4}\text{.55}\times {10}^{22}\text{moleules}$.

Explanation of Solution

When $\text{0}\text{.0755}\text{mol}$ of ammonia gas dissolves in $\text{0}\text{.155 L}$ of solution, mole concept map looks like as shown below.

Figure 1

The molecules present in $\text{1}\text{mol}$ of ${\text{NH}}_3$ gas at STP are $\text{6}\text{.02}\times {10}^{23}\text{moleules}$.

The formula to calculate the molecules occupied by $\text{0}\text{.0755}\text{mol}$ of ${\text{NH}}_3$ gas is shown below.

$\text{Number}\text{of}\text{molecules}= \text{0}\text{.0755}\text{mol}{\text{NH}}_3\times \frac{\text{Molecules}\text{in}\text{1}\text{mol}{\text{NH}}_3}{\text{1}\text{mol}{\text{NH}}_3}$

Substitute the molecules in $\text{1}\text{mol}$ of ${\text{NH}}_3$ gas in the above equation.

$\begin{array}{rll}\text{Number}\text{of}\text{molecules}& =& \text{0}\text{.0755}\text{mol}{\text{NH}}_3\times \frac{\text{6}\text{.02}\times {10}^{23}\text{moleules}}{\text{1}\text{mol}{\text{NH}}_3}\\ & =& \text{4}\text{.55}\times {10}^{22}\text{moleules}\end{array}$

Therefore, the molecules of ${\text{NH}}_3$ gas dissolved in the solution is $\text{4}\text{.55}\times {10}^{22}\text{moleules}$.

Conclusion

The molecules of ${\text{NH}}_3$ gas is $\text{4}\text{.55}\times {10}^{22}\text{moleules}$.

Interpretation Introduction

(d)

Interpretation:

A concept map is to be drawn and the molar concentration of the ammonia solution when $\text{0}\text{.0755}\text{mol}$ of ammonia gas dissolves in $\text{0}\text{.155 L}$ 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 16E

The concept map is shown below.

The molar concentration of the ammonia solution is $0.487M$.

Explanation of Solution

When $\text{0}\text{.0755}\text{mol}$ of ammonia gas dissolves in $\text{0}\text{.155 L}$ of solution, mole concept map looks like as shown below.

Figure 1

The number of moles in $\text{0}\text{.155 L}$ of solution is $\text{0}\text{.0755}\text{mol}$.

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).

$\begin{array}{rll}\text{Molar}\text{concentration}\text{of}{\text{NH}}_3& =& \frac{\text{0}\text{.0755}\text{mol}}{\text{0}\text{.155 L}}\\ & =& 0.487\text{mol}/\text{L}\end{array}$

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

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

The unit factors are given below.

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

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

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

Therefore, $0.487\text{mol}/\text{L}$ can be written as shown below.

$\begin{array}{rll}\text{Molarity}& =& 0.487\text{mol}/\text{L}\times \frac{1M}{1\text{mol}/\text{L}}\\ & =& 0.487M\end{array}$

Therefore, the molar concentration of ${\text{NH}}_3$ solution is $0.487M$.

Conclusion

The molar concentration of ${\text{NH}}_3$ solution is $0.487M$.