Chapter 4, Problem 4.14P

How many moles and how many ions of each type are present in each of the following?

1. 130. mL of 0.45 M aluminum chloride
2. 9.80 mL of a solution containing 2.59 g lithium sulfate/L
3. 245 mL of a solution containing $\text{3}\text{.68}\times {10}^{22}\text{ formula units of potassium bromide per liter}$

Interpretation Introduction

Interpretation: Number of moles and number of ions of each type in $130\text{ mL}$ of $0.45M$ aluminum chloride is to be calculated.

Concept introduction: Ionic compounds represent substance that is composed of charged ions. They are kept together by electrostatic forces. Ionic substances and electrolytes such as acid or base release ions if they are dissolved in water. These ions get separated. Positive ions of ionic compound get attracted towards negative part of water and vice-versa.

Molarity is one of the most commonly used concentration terms to determine concentration of any species. The expression for molarity of solution is as follows:

  $\text{Molarity of solution}=\frac{\text{Amount of solute}}{\text{Volume }\left(\text{L}\right)\text{ of solution}}$

Answer to Problem 4.14P

Number of molesof ${\text{Al}}^{3+}$ and ${\text{Cl}}^{-}$ is $0.0585\text{mol}$ and $0.1755\text{mol}$ respectively. Number of ${\text{Al}}^{3+}$ and ${\text{Cl}}^{-}$ ions is $3.5\times {10}^{22}$ and $1.06\times {10}^{23}$ respectively.

Explanation of Solution

Expression to calculate moles of ${\text{AlCl}}_3$ is as follows:

  $\text{Moles}=\left[\left(\text{volume of solution}\left(\text{L}\right)\right)\left(\text{molarity}\right)\right]$

Volume of solution is $130\text{ mL}$ .

Molarity of ${\text{AlCl}}_3$ is $0.45\text{ M}$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{Moles}=\left[\left(130\text{ mL}\right)\left(0.45\text{ M}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\right]\\ =0.0585\text{mol}\end{array}$

Dissociation reaction of aluminum chloride is as follows:

  ${\text{AlCl}}_3\left(s\right)\to {\text{Al}}^{3+}\left(aq\right)+3{\text{Cl}}^{-}\left(aq\right)$

According to reaction 1 mole of ${\text{AlCl}}_3$ gives 1 mole of ${\text{Al}}^{3+}$ ion and 3 moles of ${\text{Cl}}^{-}$ ion. Thus total moles of ${\text{Al}}^{3+}$ ion produced by $0.0585{\text{mol AlCl}}_3$ can be calculated as follows:

  $\begin{array}{c}{\text{Moles of Al}}^{3+}=\left(0.0585{\text{mol AlCl}}_3\right)\left(\frac{1{\text{mol Al}}^{3+}}{1{\text{mol AlCl}}_3}\right)\\ =0.0585{\text{mol Al}}^{3+}\end{array}$

Total moles of ${\text{Cl}}^{-}$ ion produced by $0.0585{\text{mol AlCl}}_3$ can be calculated as follows:

  $\begin{array}{c}{\text{Moles of Cl}}^{-}=\left(0.0585{\text{mol AlCl}}_3\right)\left(\frac{3{\text{mol Cl}}^{-}}{1{\text{mol AlCl}}_3}\right)\\ =0.1755{\text{mol Cl}}^{-}\end{array}$

Numbers of ${\text{Al}}^{3+}$ ion present in $0.0585{\text{mol Al}}^{3+}$ ions can be calculated as follows:

  $\begin{array}{c}\text{Number}{\text{of Al}}^{3+}\text{ions}=\left(0.0585{\text{mol of Al}}^{3+}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ Al}}^{3+}}{1{\text{mole of Al}}^{3+}}\right)\\ =3.522\times {10}^{22}{\text{ Al}}^{3+}\\ \approx 3.5\times {10}^{22}{\text{ Al}}^{3+}\end{array}$

Numbers of ${\text{Cl}}^{-}$ ion present in $0.1755{\text{mol Cl}}^{-}$ ions can be calculated as follows:

  $\begin{array}{c}\text{Number}{\text{of Cl}}^{-}\text{ions}=\left(0.1755{\text{mol of Cl}}^{-}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ Cl}}^{-}}{1\text{mol of}{\text{Cl}}^{-}}\right)\\ =1.06\times {10}^{23}{\text{ Cl}}^{-}\end{array}$

Interpretation Introduction

Interpretation: Number of moles and number of ions of each type in $9.80\text{ mL}$ of $2.59\text{ g lithium sulfate/L}$ solution is to be calculated.

Concept introduction:Ionic compounds represent substance that is composed of charged ions. They are kept together by electrostatic forces. Ionic substances and electrolytes such as acid or base release ions if they are dissolved in water. These ions get separated. Positive ions of ionic compound get attracted towards negative part of water and vice-versa.

Molarity is one of the most commonly used concentration terms to determine concentration of any species. The expression for molarity of solution is as follows:

  $\text{Molarity of solution}=\frac{\text{Amount of solute}}{\text{Volume }\left(\text{L}\right)\text{ of solution}}$

Answer to Problem 4.14P

Number of moles of ${\text{Li}}^{+}$ and ${\text{SO}}_4^{2-}$ is $4.62\times {10}^{-4}\text{mol}$ and $\text{2}{\text{.31×10}}^{-\text{4}}\text{ mol}$ respectively. Number of ${\text{Li}}^{+}$ and ${\text{SO}}_4^{2-}$ ions is $2.78\times {10}^{20}$ and $1.39\times {10}^{20}$ respectively.

Explanation of Solution

Expression to calculate mass of ${\text{Li}}_2{\text{SO}}_4$ is as follows:

  $\text{Mass}=\left[\left(\text{volume of solution}\left(\text{L}\right)\right)\left(\text{density}\right)\right]$

Volume of solution is $9.80\text{ mL}$ .

Density of ${\text{Li}}_2{\text{SO}}_4$ is $2.59{\text{ g Li}}_2{\text{SO}}_4\text{/L}$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{Mass}=\left[\left(9.80\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\left(2.59{\text{ g Li}}_2{\text{SO}}_4\text{/L}\right)\right]\\ =2.54\times {10}^{-2}{\text{g Li}}_2{\text{SO}}_4\end{array}$

Moles of ${\text{Li}}_2{\text{SO}}_4$ can be calculated as follows:

  $\begin{array}{c}\text{Moles=}\frac{\text{Given mass mass}}{\text{109}\text{.94 g/mol}}\\ \text{=}\frac{2.54\times {10}^{-2}\text{g}}{\text{109}\text{.94 g/mol}}\\ \text{=2}{\text{.31×10}}^{-\text{4}}\text{ mol}\end{array}$

Dissociation reaction of lithiumsulfate is as follows:

  ${\text{Li}}_2{\text{SO}}_4\left(s\right)\to 2{\text{Li}}^{+}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)$

According to reaction 1 mole of ${\text{Li}}_2{\text{SO}}_4$ gives 2 moles of ${\text{Li}}^{+}$ ion and 1 mole of ${\text{SO}}_4^{2-}$ ion. Thus total moles of ${\text{Li}}^{+}$ ion produced by $\text{2}{\text{.31×10}}^{-\text{4}}{\text{ mol Li}}_2{\text{SO}}_4$ can be calculated as follows:

  $\begin{array}{c}{\text{Moles of Li}}^{+}=\left(\text{2}{\text{.31×10}}^{-\text{4}}{\text{ mol Li}}_2{\text{SO}}_4\right)\left(\frac{2{\text{mol Li}}^{+}}{1{\text{mol Li}}_2{\text{SO}}_4}\right)\\ =4.62\times {10}^{-4}{\text{mol Li}}^{+}\end{array}$

Total moles of ${\text{SO}}_4^{2-}$ ion produced by $\text{2}{\text{.31×10}}^{-\text{4}}{\text{ mol Li}}_2{\text{SO}}_4$ can be calculated as follows:

  $\begin{array}{c}{\text{Moles of SO}}_4^{2-}=\left(\text{2}{\text{.31×10}}^{-\text{4}}{\text{ mol Li}}_2{\text{SO}}_4\right)\left(\frac{1{\text{mol SO}}_4^{2-}}{1{\text{mol Li}}_2{\text{SO}}_4}\right)\\ =\text{2}{\text{.31×10}}^{-\text{4}}{\text{ mol SO}}_4^{2-}\end{array}$

Numbers of ${\text{Li}}^{+}$ ion present in $4.62\times {10}^{-4}{\text{mol Li}}^{+}$ ions can be calculated as follows:

  $\begin{array}{c}\text{Number}{\text{of Li}}^{+}\text{ions}=\left(4.62\times {10}^{-4}{\text{mol Li}}^{+}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ Li}}^{+}}{1{\text{mole of Li}}^{+}}\right)\\ =2.78\times {10}^{20}{\text{ Li}}^{+}\end{array}$

Numbers of ${\text{SO}}_4^{2-}$ ion present in $\text{2}{\text{.31×10}}^{-\text{4}}{\text{ mol SO}}_4^{2-}$ ions can be calculated as follows:

  $\begin{array}{c}\text{Number}{\text{of SO}}_4^{2-}\text{ions}=\left(\text{2}{\text{.31×10}}^{-\text{4}}{\text{ mol SO}}_4^{2-}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ SO}}_4^{2-}}{1{\text{mol of SO}}_4^{2-}}\right)\\ =1.39\times {10}^{20}{\text{ SO}}_4^{2-}\end{array}$

Interpretation Introduction

Interpretation: Number of moles and number of ions of each type in $\text{245 mL}$ of $3.68\times {10}^{22}\text{ KBr formula unit/L}$ solution is to be calculated.

Concept introduction:Ionic compounds represent substance that is composed of charged ions. They are kept together by electrostatic forces. Ionic substances and electrolytes such as acid or base release ions if they are dissolved in water. These ions get separated. Positive ions of ionic compound get attracted towards negative part of water and vice-versa.

Molarity is one of the most commonly used concentration terms to determine concentration of any species. The expression for molarity of solution is as follows:

  $\text{Molarity of solution}=\frac{\text{Amount of solute}}{\text{Volume }\left(\text{L}\right)\text{ of solution}}$

Answer to Problem 4.14P

Number of moles of ${\text{K}}^{+}$ and ${\text{Br}}^{-}$ is $1.50\times {10}^{-2}\text{ mol}$ and $1.50\times {10}^{-2}\text{ mol}$ respectively. Number of ${\text{K}}^{+}$ and ${\text{Br}}^{-}$ ions is $9.016\times {10}^{21}$ and $9.016\times {10}^{21}$ respectively.

Explanation of Solution

Total number of formula unit of $\text{KBr}$ present in solution can be calculated as follows:

  $\begin{array}{c}\text{Formula units}\left(\text{FU}\right)=\left[\left(\text{245 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\left(3.68\times {10}^{22}\text{ KBr FU/L}\right)\right]\\ =9.016\times {10}^{21}\text{FU KBr}\end{array}$

Dissociation reaction of $\text{KBr}$ is as follows:

  $\text{KBr}\left(s\right)\to {\text{K}}^{+}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)$

According to reaction 1 formula unit of $\text{KBr}$ gives 1 mole of ${\text{K}}^{+}$ ion and 1 mole of ${\text{Br}}^{-}$ ion. Therefore, both ${\text{K}}^{+}$ and ${\text{Br}}^{-}$ ions produced by $9.016\times {10}^{21}$ formula units of $\text{KBr}$ are

  $9.016\times {10}^{21}$ .

Total moles of ${\text{K}}^{+}$ or ${\text{Br}}^{-}$ ion present in $9.016\times {10}^{21}$ formula units can be calculated as follows:

  $\begin{array}{c}\text{Moles}=\left(\left(\text{given formula unit}\right)\left(\frac{\text{1 mole}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{ formula unit}}\right)\right)\\ =\left(\left(9.016\times {10}^{21}\right)\left(\frac{\text{1 mole}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{ formula unit}}\right)\right)\\ =1.50\times {10}^{-2}\text{ mol}\end{array}$