Chapter 6, Problem 139QP

(A)

Interpretation Introduction

Interpretation:

The number of moles of ions in $2.5\text{ mol of Mg}{\left({\text{NO}}_3\right)}_2$ and the incorrect number of moles of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ that dissolve in $2.5\text{ mol of Mg}{\left({\text{NO}}_3\right)}_2$ are to be determined.

Explanation of Solution

The given balanced equation shows that one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ dissolves and dissociates into ${\text{Mg}}^{2+}$ and $2{\text{ NO}}_3^{-}$ ions, respectively. Thus, a total of three moles of ions are present in one mole of the $\text{Mg}{\left({\text{NO}}_3\right)}_2$ that dissolved. Therefore, the number of moles of ions in one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is calculated as follows.

$\begin{array}{c}\text{ Total moles of ions}=\text{ 2}\text{.5 }\cancel{\text{mol Mg}{\left({\text{NO}}_3\right)}_2}\times \frac{\text{3 moles of ions}}{\text{1 }\cancel{\text{mol Mg}{\left({\text{NO}}_3\right)}_2}}\\ =\text{7}\text{.5 moles ions}\end{array}$

Therefore, the calculated moles of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is matched with $\text{7}\text{.5 moles Mg}{\left(\text{NO}\right)}_2$ .

Three moles of ions are dissolved in one mole $\text{Mg}{\left({\text{NO}}_3\right)}_2$ . Therefore, the $2.5\text{ moles}$ of ions dissolved in one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is calculated as:

$\begin{array}{c}\text{ moles of ions}=\frac{1\text{ mol Mg}{\left({\text{NO}}_3\right)}_2}{\text{3 moles ions}}\times \text{2}\text{.5 moles ions}\\ =0.83\text{ moles Mg}{\left({\text{NO}}_3\right)}_2\end{array}$

(B)

Interpretation Introduction

Interpretation:

The number of moles of ions in $2.5\text{ mol of Mg}{\left({\text{NO}}_3\right)}_2$ is to be determined and the incorrect number of moles of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ that dissolve in $\text{5}\text{.0 mol of Mg}{\left({\text{NO}}_3\right)}_2$ are to be determined.

Explanation of Solution

The given balanced equation shows that one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ dissolves and dissociates into ${\text{Mg}}^{2+}$ and $2{\text{ NO}}_3^{-}$ ions, respectively. Thus, a total of three moles of ions are present in one mole of the $\text{Mg}{\left({\text{NO}}_3\right)}_2$ that dissolved. Therefore, the number of moles of ions in one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is calculated as follows.

$\begin{array}{c}\text{ Total moles of ions}=\text{ 2}\text{.5 }\cancel{\text{mol Mg}{\left({\text{NO}}_3\right)}_2}\times \frac{\text{3 moles of ions}}{\text{1 }\cancel{\text{mol Mg}{\left({\text{NO}}_3\right)}_2}}\\ =\text{7}\text{.5 moles ions}\end{array}$

Therefore, the calculated moles of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is matched with $\text{7}\text{.5 moles Mg}{\left(\text{NO}\right)}_2$ .

Three moles of ions are dissolved in one mole $\text{Mg}{\left({\text{NO}}_3\right)}_2$ . Therefore, the $\text{5}\text{.0 mol }$ of ions dissolved in one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is calculated as:

$\begin{array}{c}\text{ moles of ions}=\frac{1\text{ mol Mg}{\left({\text{NO}}_3\right)}_2}{\text{3 moles ions}}\times \text{5}\text{.0 moles ions }\\ =1.7\text{ moles Mg}{\left({\text{NO}}_3\right)}_2\end{array}$

(C)

Interpretation Introduction

Interpretation:

The number of moles of ions that dissolve in $2.5\text{ mol of Mg}{\left({\text{NO}}_3\right)}_2$ is to be determined.

Explanation of Solution

The given balanced equation shows that one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ dissolves and dissociates into ${\text{Mg}}^{2+}$ and $2{\text{ NO}}_3^{-}$ ions, respectively. Thus, a total of three moles of ions are present in one mole of the $\text{Mg}{\left({\text{NO}}_3\right)}_2$ that dissolved. Therefore, the number of moles of ions in one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is calculated as follows.

$\begin{array}{c}\text{ Total moles of ions}=\text{ 2}\text{.5 }\cancel{\text{mol Mg}{\left({\text{NO}}_3\right)}_2}\times \frac{\text{3 moles of ions}}{\text{1 }\cancel{\text{mol Mg}{\left({\text{NO}}_3\right)}_2}}\\ =\text{7}\text{.5 moles ions}\end{array}$

Therefore, the calculated moles of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is matched with $\text{7}\text{.5 moles Mg}{\left(\text{NO}\right)}_2$ . Hence, this is the correct option.

(D)

Interpretation Introduction

Interpretation:

The number of moles of ions in $2.5\text{ mol of Mg}{\left({\text{NO}}_3\right)}_2$ and the incorrect number of moles of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ that dissolve in $\text{20 mol of Mg}{\left({\text{NO}}_3\right)}_2$ are to be determined.

Explanation of Solution

The given balanced equation shows that one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ dissolves and dissociates into ${\text{Mg}}^{2+}$ and $2{\text{ NO}}_3^{-}$ ions, respectively. Thus, a total of three moles of ions are present in one mole of the $\text{Mg}{\left({\text{NO}}_3\right)}_2$ that dissolved. Therefore, the number of moles of ions in one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is calculated as follows.

$\begin{array}{c}\text{ Total moles of ions}=\text{ 2}\text{.5 }\cancel{\text{mol Mg}{\left({\text{NO}}_3\right)}_2}\times \frac{\text{3 moles of ions}}{\text{1 }\cancel{\text{mol Mg}{\left({\text{NO}}_3\right)}_2}}\\ =\text{7}\text{.5 moles ions}\end{array}$

Therefore, the calculated moles of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is matched with $\text{7}\text{.5 moles Mg}{\left(\text{NO}\right)}_2$ .

Three moles of ions are dissolved in one mole $\text{Mg}{\left({\text{NO}}_3\right)}_2$ . Therefore, the $\text{20 mol }$ of ions dissolved in one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is calculated as:

$\begin{array}{c}\text{ moles of ions}=\frac{1\text{ mol Mg}{\left({\text{NO}}_3\right)}_2}{\text{3 moles ions}}\times 20\text{ moles ions }\\ =6.7\text{ mol Mg}{\left({\text{NO}}_3\right)}_2\end{array}$

(E)

Interpretation Introduction

Interpretation:

The number of moles of ions in $2.5\text{ mol of Mg}{\left({\text{NO}}_3\right)}_2$ and the incorrect number of moles of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ that dissolve in $\text{0}\text{.83 mol of Mg}{\left({\text{NO}}_3\right)}_2$ are to be determined.

Explanation of Solution

The given balanced equation shows that one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ dissolves and dissociates into ${\text{Mg}}^{2+}$ and $2{\text{ NO}}_3^{-}$ ions, respectively. Thus, a total of three moles of ions are present in one mole of the $\text{Mg}{\left({\text{NO}}_3\right)}_2$ that dissolved. Therefore, the number of moles of ions in one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is calculated as follows.

$\begin{array}{c}\text{ Total moles of ions}=\text{ 2}\text{.5 }\cancel{\text{mol Mg}{\left({\text{NO}}_3\right)}_2}\times \frac{\text{3 moles of ions}}{\text{1 }\cancel{\text{mol Mg}{\left({\text{NO}}_3\right)}_2}}\\ =\text{7}\text{.5 moles ions}\end{array}$

Therefore, the calculated moles of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is matched with $\text{7}\text{.5 moles Mg}{\left(\text{NO}\right)}_2$ .

Three moles of ions are dissolved in one mole $\text{Mg}{\left({\text{NO}}_3\right)}_2$ . Therefore, the $\text{0}\text{.83 mol }$ of ions dissolved in one mole of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is calculated as:

$\begin{array}{c}\text{mol of ions}=\frac{1\text{ mol Mg}{\left({\text{NO}}_3\right)}_2}{\text{3 moles ions}}\times 0.83\text{ moles ions }\\ =0.28\text{ mol Mg}{\left({\text{NO}}_3\right)}_2\end{array}$