Chapter 4, Problem 4.15P

a)

Interpretation Introduction

Interpretation: Number of moles and number of ions of each type in $88\text{ mL}$ of $1.75M$ magnesium 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 4.15P

Number of molesof ${\text{Mg}}^{2+}$ and ${\text{Cl}}^{-}$ is $0.154\text{mol}$ and $0.308\text{mol}$ respectively. Number of ${\text{Mg}}^{2+}$ and ${\text{Cl}}^{-}$ ions is $9.27\times {10}^{22}$ and $1.85\times {10}^{23}$ respectively.

Explanation of Solution

Expression to calculate moles of ${\text{MgCl}}_2$ 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 $88\text{ mL}$ .

Molarity of ${\text{MgCl}}_2$ is $1.75M$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{Moles}=\left[\left(88\text{ mL}\right)\left(1.75M\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\right]\\ =0.154\text{mol}\end{array}$

Dissociation reaction of magnesium chloride is as follows:

  ${\text{MgCl}}_2\left(s\right)\to {\text{Mg}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$

According to reaction, 1 mole of ${\text{MgCl}}_2$ gives 1 mole of ${\text{Mg}}^{2+}$ ion and 2 moles of ${\text{Cl}}^{-}$ ion. Thus moles of ${\text{Mg}}^{2+}$ ion produced by $0.154{\text{mol MgCl}}_2$ can be calculated as follows:

  $\begin{array}{c}{\text{Moles of Mg}}^{2+}=\left(0.154{\text{mol MgCl}}_2\right)\left(\frac{1{\text{mol Mg}}^{2+}}{1{\text{mol MgCl}}_2}\right)\\ =0.154{\text{mol Mg}}^{2+}\end{array}$

Moles of ${\text{Cl}}^{-}$ ion produced by $0.154{\text{mol MgCl}}_2$ can be calculated as follows:

  $\begin{array}{c}{\text{Moles of Cl}}^{-}=\left(0.154{\text{mol MgCl}}_2\right)\left(\frac{2{\text{mol Cl}}^{-}}{1{\text{mol MgCl}}_2}\right)\\ =0.308{\text{mol Cl}}^{-}\end{array}$

Numbers of ${\text{Mg}}^{2+}$ ion present in $0.154{\text{mol Mg}}^{2+}$ ions can be calculated as follows:

  $\begin{array}{c}\text{Number}{\text{of Mg}}^{2+}\text{ions}=\left(0.154{\text{mol Mg}}^{2+}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ Mg}}^{2+}}{1{\text{mole of Mg}}^{2+}}\right)\\ =9.27\times {10}^{22}{\text{ Mg}}^{2+}\end{array}$

Numbers of ${\text{Cl}}^{-}$ ion present in $0.308{\text{mol Cl}}^{-}$ ions can be calculated as follows:

  $\begin{array}{c}\text{Number}{\text{of Cl}}^{-}\text{ions}=\left(0.308{\text{mol Cl}}^{-}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ Cl}}^{-}}{1\text{mol of}{\text{Cl}}^{-}}\right)\\ =1.85\times {10}^{23}{\text{ Cl}}^{-}\end{array}$

b)

Interpretation Introduction

Interpretation: Number of moles and number of ions of each type in $321\text{ mL}$ of $0.22\text{ g aluminium sulfate/L}$ 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 4.15P

Number of molesof ${\text{Al}}^{3+}$ and ${\text{SO}}_4^{2-}$ is $\text{4}{\text{.12×10}}^{-\text{4}}\text{ mol}$ and $\text{6}{\text{.18×10}}^{-\text{4}}\text{ mol}$ respectively. Number of ${\text{Al}}^{3+}$ and ${\text{SO}}_4^{2-}$ ions is $2.48\times {10}^{20}$ and $3.72\times {10}^{20}$ respectively.

Explanation of Solution

Expression to calculate mass of ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3$ 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 $321\text{ mL}$ .

Density of ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3$ is $0.22{\text{ g Al}}_2{\left({\text{SO}}_4\right)}_3\text{/L}$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{Mass}=\left[\left(321\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\left(0.22{\text{ g Al}}_2{\left({\text{SO}}_4\right)}_3\text{/L}\right)\right]\\ =7.06\times {10}^{-2}{\text{g Li}}_2{\text{SO}}_4\end{array}$

Moles of ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3$ can be calculated as follows:

  $\begin{array}{c}\text{Moles=}\frac{\text{Given mass mass}}{342.14\text{ g/mol}}\\ \text{=}\frac{7.06\times {10}^{-2}\text{g}}{342.14\text{ g/mol}}\\ =\text{2}{\text{.06×10}}^{-\text{4}}\text{ mol}\end{array}$

Dissociation reaction of ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3$ is as follows:

  ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3\left(s\right)\to 2{\text{Al}}^{3+}\left(aq\right)+3{\text{SO}}_4^{2-}\left(aq\right)$

According to reaction 1 mole of ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3$ gives 2 mole of ${\text{Al}}^{3+}$ ion and 3 mole of ${\text{SO}}_4^{2-}$ ion. Thus moles of ${\text{Al}}^{3+}$ ion produced by $\text{2}{\text{.06×10}}^{-\text{4}}{\text{ mol Al}}_2{\left({\text{SO}}_4\right)}_3$ can be calculated as follows:

  $\begin{array}{c}{\text{Moles of Al}}^{3+}=\left(\text{2}{\text{.06×10}}^{-\text{4}}{\text{ mol Al}}_2{\left({\text{SO}}_4\right)}_3\right)\left(\frac{2{\text{mol Al}}^{3+}}{1{\text{mol Al}}_2{\left({\text{SO}}_4\right)}_3}\right)\\ =\text{4}{\text{.12×10}}^{-\text{4}}{\text{ mol Al}}^{3+}\end{array}$

Moles of ${\text{SO}}_4^{2-}$ ion produced by $\text{2}{\text{.06×10}}^{-\text{4}}{\text{ mol Al}}_2{\left({\text{SO}}_4\right)}_3$ can be calculated as follows:

  $\begin{array}{c}{\text{Moles of SO}}_4^{2-}=\left(\text{2}{\text{.06×10}}^{-\text{4}}{\text{ mol Al}}_2{\left({\text{SO}}_4\right)}_3\right)\left(\frac{3{\text{mol SO}}_4^{2-}}{1{\text{mol Al}}_2{\left({\text{SO}}_4\right)}_3}\right)\\ =\text{6}{\text{.18×10}}^{-\text{4}}{\text{ mol SO}}_4^{2-}\end{array}$

Numbers of ${\text{Al}}^{3+}$ ion present in $\text{4}{\text{.12×10}}^{-\text{4}}{\text{ mol Al}}^{3+}$ ions can be calculated as follows:

  $\begin{array}{c}\text{Number}{\text{of Al}}^{3+}\text{ions}=\left(\text{4}{\text{.12×10}}^{-\text{4}}{\text{ mol Al}}^{3+}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{Al}}^{3+}}{1{\text{mole of Al}}^{3+}}\right)\\ =2.48\times {10}^{20}{\text{ Al}}^{3+}\end{array}$

Numbers of ${\text{SO}}_4^{2-}$ ion present in $\text{6}{\text{.18×10}}^{-\text{4}}{\text{ mol SO}}_4^{2-}$ ions can be calculated as follows:

  $\begin{array}{c}\text{Number}{\text{of SO}}_4^{2-}\text{ions}=\left(\text{6}{\text{.18×10}}^{-\text{4}}{\text{ mol SO}}_4^{2-}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ SO}}_4^{2-}}{1{\text{mol of SO}}_4^{2-}}\right)\\ =3.72\times {10}^{20}{\text{ SO}}_4^{2-}\end{array}$

c)

Interpretation Introduction

Interpretation: Number of moles and number of ions of each type in $1.65\text{ mL}$ of $8.83\times {10}^{21}{\text{ CsNO}}_3\text{ formula unit/L}$ 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 4.15P

Number of molesof ${\text{Cs}}^{+}$ and ${\text{NO}}_3^{-}$ is $2.42\times {10}^{-5}\text{ mol}$ and $2.42\times {10}^{-5}\text{ mol}$ respectively. Number of ${\text{Cs}}^{+}$ and ${\text{NO}}_3^{-}$ ions is $1.46\times {10}^{19}$ and $1.46\times {10}^{19}$ respectively.

Explanation of Solution

Total number of formula unit of ${\text{CsNO}}_3$ present in solution can be calculated as follows:

  $\begin{array}{c}\text{Formula units}\left(\text{FU}\right)=\left[\left(1.65\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\left(8.83\times {10}^{21}{\text{ CsNO}}_3\text{ FU/L}\right)\right]\\ =1.46\times {10}^{19}{\text{FU CsNO}}_3\end{array}$

Dissociation reaction of ${\text{CsNO}}_3$ is as follows:

  ${\text{CsNO}}_3\left(s\right)\to {\text{Cs}}^{+}\left(aq\right)+{\text{NO}}_3^{-}\left(aq\right)$

According to reaction 1 formula unit of ${\text{CsNO}}_3$ gives 1 mole of ${\text{Cs}}^{+}$ ion and 1 mole of ${\text{NO}}_3^{-}$ ion. Therefore, both ${\text{Cs}}^{+}$ and ${\text{NO}}_3^{-}$ ions produced by $1.46\times {10}^{19}$ formula units of ${\text{CsNO}}_3$ are $1.46\times {10}^{19}$ .

Moles of ${\text{Cs}}^{+}$ or ${\text{NO}}_3^{-}$ ion present in $1.46\times {10}^{19}$ formula units can be calculated as follows:

  $\begin{array}{c}\text{Moles}=\left(\left(\text{given formula unit}\right)\left(\frac{\text{1 mole}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{ formula unit}}\right)\right)\\ =\left(\left(1.46\times {10}^{19}\right)\left(\frac{\text{1 mole}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{ formula unit}}\right)\right)\\ =2.42\times {10}^{-5}\text{ mol}\end{array}$