Chapter 4, Problem 4.26P

(a)

Interpretation Introduction

Interpretation:

The number of moles and number of ions of each type in $130\text{ mL}$ of $0.45M$ aluminium chloride is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol/L}$.

The expression to calculate the moles of the compound when molarity of solution and volume of solution are given is as follows:

$\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{volume of solution}\left(\text{L}\right)\\ \left(\frac{\text{molarity}\text{of}\text{solution}\left(\text{mol}\right)}{\text{1L of solution}}\right)\end{array}\right]$

The expression to calculate the moles of ions is as follows:

$\text{moles of}\text{ion of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\left(\text{moles of compound}\left(\text{mol}\right)\right)\\ \left(\frac{\text{total moles of ion}\left(\text{mol}\right)}{1\text{mole of compound}}\right)\end{array}\right]$

The expression to calculate the number of ions is as follows:

$\text{number}\text{of ions}=\left(\text{moles of ions}\left(\text{mol}\right)\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}\text{ions}}{1\text{mole of ions}}\right)$

Answer to Problem 4.26P

The number of moles of is ${\text{Al}}^{3+}$ and ${\text{Cl}}^{-}$ is $0.058\text{mol}$ and $0.18\text{mol}$ respectively. The number of ions ${\text{Al}}^{3+}$ and ${\text{Cl}}^{-}$ is $3.5\times {10}^{22}$ and $1.1\times {10}^{23}$ respectively.

Explanation of Solution

The expression to calculate the moles of aluminium chloride is:

$\text{Moles of aluminium chloride}\left(\text{mol}\right)=\left[\begin{array}{c}\text{volume of solution}\left(\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ \left(\frac{\text{molarity}\text{of}\text{aluminium chloride}\left(\text{mol}\right)}{\text{1L of solution}}\right)\end{array}\right]$

The relation between $\text{mL}$ and $\text{L}$ is:

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

Substitute $130\text{ mL}$ for the volume of solution and $0.45\text{mol/L}$ for molarity of aluminium chloride in the above equation as follows:

$\begin{array}{c}\text{Moles of aluminium chloride}\left(\text{mol}\right)=130\text{ mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(\frac{0.45\text{mol}}{\text{1L of solution}}\right)\\ =0.0585\text{mol}\end{array}$

One mole of aluminium chloride $\left({\text{AlCl}}_3\right)$ on dissociation produces one mole of ${\text{Al}}^{3+}$ ion and three moles of ${\text{Cl}}^{-}$ ion.

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

The expression to calculate the amount of ${\text{Al}}^{3+}$ ion in moles is as follows:

$\text{amount}\text{of}{\text{Al}}^{3+}\left(\text{mol}\right)=\left({\text{moles of AlCl}}_3\left(\text{mol}\right)\right)\left(\frac{{\text{moles of Al}}^{3+}\text{ion}\left(\text{mol}\right)}{1{\text{mole of AlCl}}_3}\right)$

Substitute $0.0585\text{mol}$ for moles of ${\text{AlCl}}_3$ and $\text{1 mol}$ for moles of ${\text{Al}}^{3+}$ ion in the above equation as follows:

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

The expression to calculate the number of ${\text{Al}}^{3+}$ ions is:

$\text{number}{\text{of Al}}^{3+}\text{ions}=\left({\text{moles of Al}}^{3+}\left(\text{mol}\right)\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{Al}}^{3+}\text{ions}}{1{\text{mole of Al}}^{3+}}\right)$

Substitute $0.0585\text{mol}$ for moles of ${\text{Al}}^{3+}$ in the above equation as follows:

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

The expression to calculate the amount of ${\text{Cl}}^{-}$ ion in moles is as follows:

$\text{amount}\text{of}{\text{Cl}}^{-}\left(\text{mol}\right)=\left({\text{moles of AlCl}}_3\left(\text{mol}\right)\right)\left(\frac{{\text{moles of Cl}}^{-}\text{ion}\left(\text{mol}\right)}{1{\text{mole of AlCl}}_3}\right)$

Substitute $0.0585\text{mol}$ for moles of ${\text{AlCl}}_3$ and $\text{3 mol}$ for moles of ${\text{Cl}}^{-}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{Cl}}^{-}\left(\text{mol}\right)=\left(0.0585\text{mol}\right)\left(\frac{\text{3 mol}}{1{\text{mole of AlCl}}_3}\right)\\ =0.18\text{mol}\end{array}$

The expression to calculate the number of ${\text{Cl}}^{-}$ ions is:

$\text{number}{\text{of Cl}}^{-}\text{ions}=\left({\text{moles of Cl}}^{-}\left(\text{mol}\right)\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{Cl}}^{-}\text{ions}}{1{\text{mole of Cl}}^{-}}\right)$

Substitute $0.0585\text{mol}$ for moles of ${\text{Cl}}^{-}$ in the above equation as follows:

$\begin{array}{c}\text{number}{\text{of Cl}}^{-}\text{ions}=\left(0.18\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{Cl}}^{-}\text{ions}}{1{\text{mole of Cl}}^{-}}\right)\\ =1.0568\times {10}^{23}\\ \approx 1.1\times {10}^{23}\end{array}$

Conclusion

The number of moles of is ${\text{Al}}^{3+}$ and ${\text{Cl}}^{-}$ is $0.058\text{mol}$ and $0.18\text{mol}$ respectively. The number of ions ${\text{Al}}^{3+}$ and ${\text{Cl}}^{-}$ is $3.5\times {10}^{22}$ and $1.1\times {10}^{23}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The number of moles and number of ions of each type in $9.80\text{ mL}$ of a solution containing $2.59\text{ g}$ lithium sulphate per litre is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol/L}$.

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

$\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of compound}\left(\text{g}\right)\\ \left(\frac{\text{1}\text{mole}\text{of compound}\left(\text{mol}\right)}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$

The expression to calculate the moles of ions is as follows:

$\text{moles of}\text{ion of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\left(\text{moles of compound}\left(\text{mol}\right)\right)\\ \left(\frac{\text{total moles of ion}\left(\text{mol}\right)}{1\text{mole of compound}}\right)\end{array}\right]$

The expression to calculate the number of ions is as follows:

$\text{number}\text{of ions}=\left(\text{moles of ions}\left(\text{mol}\right)\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}\text{ions}}{1\text{mole of ions}}\right)$

Answer to Problem 4.26P

The number of moles of is ${\text{Li}}^{+}$ and ${\text{SO}}_4^{2-}$ is $4.62\times {10}^{-4}\text{mol}$ and $2.31\times {10}^{-4}\text{mol}$ respectively. The number of ions ${\text{Li}}^{+}$ and ${\text{SO}}_4^{2-}$ is $2.78\times {10}^{20}$ and $1.39\times {10}^{20}$ respectively.

Explanation of Solution

The molecular mass of lithium sulphate $\left({\text{Li}}_2{\text{SO}}_4\right)$ is $109.94\text{ g/mol}$.

The expression to calculate the moles of lithium sulphate $\left({\text{Li}}_2{\text{SO}}_4\right)$ per liter is:

${\text{Moles of Li}}_2{\text{SO}}_4\left(\text{mol}\right)=\left[\begin{array}{c}\text{volume of solution}\left(\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ \left(\frac{\text{given mass}{\text{of Li}}_2{\text{SO}}_4\left(\text{g}\right)}{1\text{L}}\right)\\ \left(\frac{\text{1}\text{mole}{\text{of Li}}_2{\text{SO}}_4\left(\text{mol}\right)}{{\text{molecular mass of Li}}_2{\text{SO}}_4\left(\text{g}\right)}\right)\end{array}\right]$

Substitute $9.80\text{ mL}$ for the volume of solution, $109.94\text{ g/mol}$ for the molecular mass of ${\text{Li}}_2{\text{SO}}_4$ and $2.59\text{ g}$ for given mass of ${\text{Li}}_2{\text{SO}}_4$ in the above equation as follows:

$\begin{array}{c}{\text{Moles of Li}}_2{\text{SO}}_4\left(\text{mol}\right)=\left[\begin{array}{c}9.80\text{ mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ \left(\frac{2.59\text{ g}}{1\text{L}}\right)\left(\frac{\text{1}\text{mole}{\text{of Li}}_2{\text{SO}}_4\left(\text{mol}\right)}{109.94\text{ g}}\right)\end{array}\right]\\ =2.3087\times {10}^{-4}\text{mol}\end{array}$

One mole of lithium sulphate $\left({\text{Li}}_2{\text{SO}}_4\right)$ on dissociation produces two moles of ${\text{Li}}^{+}$ ion and one mole of ${\text{SO}}_4^{2-}$ ion.

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

The expression to calculate the amount of ${\text{Li}}^{+}$ ion in moles is as follows:

$\text{amount}\text{of}{\text{Li}}^{+}\left(\text{mol}\right)=\left({\text{moles of Li}}_2{\text{SO}}_4\left(\text{mol}\right)\right)\left(\frac{{\text{moles of Li}}^{+}\text{ion}\left(\text{mol}\right)}{1{\text{mole of Li}}_2{\text{SO}}_4}\right)$

Substitute $2.3087\times {10}^{-4}\text{mol}$ for moles of ${\text{Li}}_2{\text{SO}}_4$ and $\text{2 mol}$ for moles of ${\text{Li}}^{+}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{Li}}^{+}\left(\text{mol}\right)=\left(2.3087\times {10}^{-4}\text{mol}\right)\left(\frac{\text{2 mol}}{1{\text{mole of Li}}_2{\text{SO}}_4}\right)\\ =4.62\times {10}^{-4}\text{mol}\end{array}$

The expression to calculate the number of ${\text{Li}}^{+}$ ions is:

$\text{number}{\text{of Li}}^{+}\text{ions}=\left({\text{moles of Li}}^{+}\left(\text{mol}\right)\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{Li}}^{+}\text{ ions}}{1{\text{mole of Li}}^{+}}\right)$

Substitute $4.62\times {10}^{-4}\text{mol}$ for moles of ${\text{Li}}^{+}$ in the above equation as follows:

$\begin{array}{c}\text{number}{\text{of Li}}^{+}\text{ions}=\left(4.62\times {10}^{-4}\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ Li}}^{+}\text{ ions}}{1{\text{mole of Li}}^{+}}\right)\\ =2.7806\times {10}^{20}\\ \approx 2.78\times {10}^{20}\end{array}$

The expression to calculate the amount of ${\text{SO}}_4^{2-}$ ion in moles is as follows:

$\text{amount}\text{of}{\text{SO}}_4^{2-}\left(\text{mol}\right)=\left({\text{moles of Li}}_2{\text{SO}}_4\left(\text{mol}\right)\right)\left(\frac{{\text{moles of SO}}_4^{2-}\text{ion}\left(\text{mol}\right)}{1{\text{mole of Li}}_2{\text{SO}}_4}\right)$

Substitute $2.3087\times {10}^{-4}\text{mol}$ for moles of ${\text{Li}}_2{\text{SO}}_4$ and $\text{1 mol}$ for moles of ${\text{SO}}_4^{2-}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{SO}}_4^{2-}\left(\text{mol}\right)=\left(2.3087\times {10}^{-4}\text{mol}\right)\left(\frac{\text{1 mol}}{1{\text{mole of Li}}_2{\text{SO}}_4}\right)\\ =2.3087\times {10}^{-4}\text{mol}\end{array}$

The expression to calculate the number of ${\text{SO}}_4^{2-}$ ions is:

$\text{number}{\text{of SO}}_4^{2-}\text{ions}=\left({\text{moles of SO}}_4^{2-}\left(\text{mol}\right)\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{SO}}_4^{2-}\text{ ions}}{1{\text{mole of SO}}_4^{2-}}\right)$

Substitute $2.3087\times {10}^{-4}\text{mol}$ for moles of ${\text{SO}}_4^{2-}$ in the above equation as follows:

$\begin{array}{c}\text{number}{\text{of SO}}_4^{2-}\text{ions}=\left(2.3087\times {10}^{-4}\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ SO}}_4^{2-}\text{ ions}}{1{\text{mole of SO}}_4^{2-}}\right)\\ =1.39030\times {10}^{20}\\ \approx 1.39\times {10}^{20}\end{array}$

Conclusion

The number of moles of is ${\text{Li}}^{+}$ and ${\text{SO}}_4^{2-}$ is $4.62\times {10}^{-4}\text{mol}$ and $2.31\times {10}^{-4}\text{mol}$ respectively. The number of ions ${\text{Li}}^{+}$ and ${\text{SO}}_4^{2-}$ is $2.78\times {10}^{20}$ and $1.39\times {10}^{20}$ respectively.

(c)

Interpretation Introduction

Interpretation:

The number of moles and number of ions of each type in $245\text{ mL}$ of a solution containing $3.68\times {10}^{22}$ formula units of potassium bromide per liter is to be calculated.

Concept introduction:

A formula unit is used for the ionic compound to represent their empirical formula. The expression to calculate the moles of a compound when the volume of solution and formula unit of a compound is given is as follows:

$\text{moles of a compound}\left(\text{mol}\right)=\left[\begin{array}{c}\left(\text{volume of solution}\left(\text{L}\right)\right)\\ \left(\text{given formula unit of compound}\left(\text{FU}\right)\right)\\ \left(\frac{\text{1 mole of compound}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{FU}}\right)\end{array}\right]$

The expression to calculate the moles of ions is as follows:

$\text{moles of}\text{ion of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\left(\text{moles of compound}\left(\text{mol}\right)\right)\\ \left(\frac{\text{total moles of ion}\left(\text{mol}\right)}{1\text{mole of compound}}\right)\end{array}\right]$

The expression to calculate the number of ions is as follows:

$\text{number}\text{of ions}=\left(\text{moles of ions}\left(\text{mol}\right)\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}\text{ions}}{1\text{mole of ions}}\right)$

Answer to Problem 4.26P

The number of moles of is ${\text{K}}^{+}$ and ${\text{Br}}^{-}$ is $1.50\times {10}^{-2}\text{mol}$ and $1.50\times {10}^{-2}\text{mol}$ respectively. The number of ions ${\text{K}}^{+}$ and ${\text{Br}}^{-}$ is $9.02\times {10}^{21}$ and $9.02\times {10}^{21}$ respectively.

Explanation of Solution

One mole of potassium bromide $\left(\text{KBr}\right)$ on dissociation produces one mole of ${\text{K}}^{+}$ ion and one mole of ${\text{Br}}^{-}$ ion.

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

The expression to calculate the moles of $\text{KBr}$ is:

$\text{moles of KBr}\left(\text{mol}\right)=\left[\begin{array}{c}\left(\text{volume of solution}\left(\text{mL}\right)\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ \left(\text{given formula unit of KBr}\left(\text{FU}\right)\right)\\ \left(\frac{\text{1 mole of KBr}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{FU}}\right)\end{array}\right]$

Substitute $3.68\times {10}^{22}$ formula unit for given formula unit of $\text{KBr}$ and $245\text{ mL}$ for the volume of solution in the above equation as follows:

$\begin{array}{c}\text{moles of KBr}\left(\text{mol}\right)=\left[\begin{array}{c}\left(245\text{ mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ \left(3.68\times {10}^{22}\right)\left(\frac{\text{1 mole of KBr}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{FU}}\right)\end{array}\right]\\ =0.01479\text{mol}\end{array}$

The expression to calculate the amount of ${\text{K}}^{+}$ ion in moles is as follows:

$\text{amount}\text{of}{\text{K}}^{+}\left(\text{mol}\right)=\left(\text{moles of KBr}\left(\text{mol}\right)\right)\left(\frac{{\text{moles of K}}^{+}\text{ion}\left(\text{mol}\right)}{1\text{mole of KBr}}\right)$

Substitute $0.01479\text{mol}$ for moles of $\text{KBr}$ and $\text{1 mol}$ for moles of ${\text{K}}^{+}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{K}}^{+}\left(\text{mol}\right)=\left(0.01479\text{mol}\right)\left(\frac{\text{1 mol}}{1\text{mole of KBr}}\right)\\ =1.50\times {10}^{-2}\text{mol}\end{array}$

The expression to calculate the number of ${\text{K}}^{+}$ ions is:

$\text{number}{\text{of K}}^{+}\text{ions}=\left({\text{moles of K}}^{+}\left(\text{mol}\right)\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{K}}^{+}\text{ ions}}{1{\text{mole of K}}^{+}}\right)$

Substitute $1.50\times {10}^{-2}\text{mol}$ for moles of ${\text{K}}^{+}$ in the above equation as follows:

$\begin{array}{c}\text{number}{\text{of K}}^{+}\text{ions}=\left(1.50\times {10}^{-2}\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ K}}^{+}\text{ ions}}{1{\text{mole of K}}^{+}}\right)\\ =9.016\times {10}^{21}\\ \approx 9.02\times {10}^{21}\end{array}$

The expression to calculate the amount of ${\text{Br}}^{-}$ ion in moles is as follows:

$\text{amount}\text{of}{\text{Br}}^{-}\left(\text{mol}\right)=\left(\text{moles of KBr}\left(\text{mol}\right)\right)\left(\frac{{\text{moles of Br}}^{-}\text{ion}\left(\text{mol}\right)}{1\text{mole of KBr}}\right)$

Substitute $0.01479\text{mol}$ for moles of $\text{KBr}$ and $\text{1 mol}$ for moles of ${\text{Br}}^{-}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{Br}}^{-}\left(\text{mol}\right)=\left(0.01479\text{mol}\right)\left(\frac{\text{1 mol}}{1\text{mole of KBr}}\right)\\ =1.50\times {10}^{-2}\text{mol}\end{array}$

The expression to calculate the number of ${\text{Br}}^{-}$ ions is:

$\text{number}{\text{of Br}}^{-}\text{ions}=\left({\text{moles of Br}}^{-}\left(\text{mol}\right)\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{Br}}^{-}\text{ ions}}{1{\text{mole of Br}}^{-}}\right)$

Substitute $1.50\times {10}^{-2}\text{mol}$ for moles of ${\text{Br}}^{-}$ in the above equation as follows:

$\begin{array}{c}\text{number}{\text{of Br}}^{-}\text{ions}=\left(1.50\times {10}^{-2}\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ Br}}^{-}\text{ ions}}{1{\text{mole of Br}}^{-}}\right)\\ =9.016\times {10}^{21}\\ \approx 9.02\times {10}^{21}\end{array}$

Conclusion

The number of moles of is ${\text{K}}^{+}$ and ${\text{Br}}^{-}$ is $1.50\times {10}^{-2}\text{mol}$ and $1.50\times {10}^{-2}\text{mol}$ respectively. The number of ions ${\text{K}}^{+}$ and ${\text{Br}}^{-}$ is $9.02\times {10}^{21}$ and $9.02\times {10}^{21}$ respectively.