Chapter 4, Problem 4.27P

(a)

Interpretation Introduction

Interpretation:

The 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:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per liter 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 4.27P

The number of moles of is ${\text{Mg}}^{2+}$ and ${\text{Cl}}^{-}$ ions is $0.15\text{mol}$ and $0.31\text{mol}$ respectively. The number of ions ${\text{Mg}}^{2+}$ and ${\text{Cl}}^{-}$ is $9.3\times {10}^{22}$ and $1.9\times {10}^{23}$ respectively.

Explanation of Solution

The expression to calculate the moles of magnesium chloride is:

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

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

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

Substitute $88\text{ mL}$ for the volume of solution and $1.75\text{ mol/L}$ for molarity of magnesium chloride in the above equation as follows:

$\begin{array}{c}\text{Moles of magnesium chloride}\left(\text{mol}\right)=88\text{ mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(\frac{1.75\text{mol}}{\text{1L of solution}}\right)\\ =0.154\text{mol}\end{array}$

One mole of magnesium chloride $\left({\text{MgCl}}_2\right)$ on dissociation produces one mole of ${\text{Mg}}^{2+}$ ion and two moles of ${\text{Cl}}^{-}$ ion.

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

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

$\text{amount}\text{of}{\text{Mg}}^{2+}\left(\text{mol}\right)=\left({\text{moles of MgCl}}_2\left(\text{mol}\right)\right)\left(\frac{{\text{moles of Mg}}^{2+}\text{ion}\left(\text{mol}\right)}{1{\text{mole of MgCl}}_2}\right)$

Substitute $0.154\text{mol}$ for moles of ${\text{MgCl}}_2$ and $\text{1 mol}$ for moles of ${\text{Mg}}^{2+}$ ion in the above equation as follows:

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

The expression to calculate the number of ${\text{Mg}}^{2+}$ ions is:

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

Substitute $0.154\text{mol}$ for moles of ${\text{Mg}}^{2+}$ in the above equation as follows:

$\begin{array}{c}\text{number}{\text{of Mg}}^{2+}\text{ions}=\left(0.154\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ Al}}^{3+}\text{ions}}{1{\text{mole of Mg}}^{2+}}\right)\\ =9.2738\times {10}^{22}\\ \approx 9.3\times {10}^{22}\end{array}$

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

$\text{amount}\text{of}{\text{Cl}}^{-}\left(\text{mol}\right)=\left({\text{moles of MgCl}}_2\left(\text{mol}\right)\right)\left(\frac{{\text{moles of Cl}}^{-}\text{ion}\left(\text{mol}\right)}{1{\text{mole of MgCl}}_2}\right)$

Substitute $0.154\text{mol}$ for moles of ${\text{MgCl}}_2$ and $\text{2 mol}$ for moles of ${\text{Cl}}^{-}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{Cl}}^{-}\left(\text{mol}\right)=\left(0.154\text{mol}\right)\left(\frac{\text{2 mol}}{1{\text{mole of MgCl}}_2}\right)\\ =0.31\text{mol}\end{array}$

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 $0.31\text{mol}$ for moles of ${\text{Cl}}^{-}$ in the above equation as follows:

$\begin{array}{c}\text{number}{\text{of Cl}}^{-}\text{ions}=\left(0.31\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{Cl}}^{-}\text{ions}}{1{\text{mole of Cl}}^{-}}\right)\\ =1.85477\times {10}^{23}\\ \approx 1.9\times {10}^{23}\end{array}$

Conclusion

The number of moles of is ${\text{Mg}}^{2+}$ and ${\text{Cl}}^{-}$ is $0.15\text{mol}$ and $0.31\text{mol}$ respectively. The number of ions ${\text{Mg}}^{2+}$ and ${\text{Cl}}^{-}$ is $9.3\times {10}^{22}$ and $1.9\times {10}^{23}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The number of moles and number of ions of each type in $321\text{ mL}$ of a solution containing $0.22\text{ g}$ aluminium sulfate per liter 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 liter 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 4.27P

The number of moles of is ${\text{Al}}^{3+}$ and ${\text{SO}}_4^{2-}$ is $4.1\times {10}^{-4}\text{mol}$ and $6.2\times {10}^{-4}\text{mol}$ respectively. The number of ions ${\text{Al}}^{3+}$ and ${\text{SO}}_4^{2-}$ is $2.5\times {10}^{20}$ and $3.7\times {10}^{20}$ respectively.

Explanation of Solution

The molecular mass of aluminium sulfate $\left({\text{Al}}_2{\left({\text{SO}}_4\right)}_3\right)$ is $342.14\text{ g/mol}$.

The expression to calculate the moles of aluminium sulfate $\left({\text{Al}}_2{\left({\text{SO}}_4\right)}_3\right)$ per liter is:

${\text{Moles of Al}}_2{\left({\text{SO}}_4\right)}_3\left(\text{mol}\right)=\left[\begin{array}{c}\text{volume of solution}\left(\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ \left(\frac{\text{given mass}{\text{of Al}}_2{\left({\text{SO}}_4\right)}_3\left(\text{g}\right)}{1\text{L}}\right)\\ \left(\frac{\text{1}\text{mole}{\text{of Al}}_2{\left({\text{SO}}_4\right)}_3\left(\text{mol}\right)}{{\text{molecular mass of Al}}_2{\left({\text{SO}}_4\right)}_3\left(\text{g}\right)}\right)\end{array}\right]$

Substitute $321\text{ mL}$ for the volume of solution, $342.14\text{ g/mol}$ for the molecular mass of ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3$ and $0.22\text{ g}$ for given mass of ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3$ in the above equation as follows:

$\begin{array}{c}{\text{Moles of Al}}_2{\left({\text{SO}}_4\right)}_3\left(\text{mol}\right)=\left[\begin{array}{c}321\text{ mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ \left(\frac{0.22\text{ g}}{1\text{L}}\right)\left(\frac{\text{1}\text{mole}{\text{of Al}}_2{\left({\text{SO}}_4\right)}_3\left(\text{mol}\right)}{109.94\text{ g}}\right)\end{array}\right]\\ =2.06406\times {10}^{-4}\text{mol}\end{array}$

One mole of aluminium sulfate $\left({\text{Al}}_2{\left({\text{SO}}_4\right)}_3\right)$ on dissociation produces two moles of ${\text{Al}}^{3+}$ ion and three moles of ${\text{SO}}_4^{2-}$ ion.

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

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

$\text{amount}\text{of}{\text{Al}}^{3+}\left(\text{mol}\right)=\left({\text{moles of Al}}_2{\left({\text{SO}}_4\right)}_3\left(\text{mol}\right)\right)\left(\frac{{\text{moles of Al}}^{3+}\text{ion}\left(\text{mol}\right)}{1{\text{mole of Al}}_2{\left({\text{SO}}_4\right)}_3}\right)$

Substitute $2.06406\times {10}^{-4}\text{mol}$ for moles of ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3$ and $\text{2 mol}$ for moles of ${\text{Al}}^{3+}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{Al}}^{3+}\left(\text{mol}\right)=\left(2.06406\times {10}^{-4}\text{mol}\right)\left(\frac{\text{2 mol}}{1{\text{mole of Al}}_2{\left({\text{SO}}_4\right)}_3}\right)\\ =4.1\times {10}^{-4}\text{mol}\end{array}$

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

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

Substitute $4.1\times {10}^{-4}\text{mol}$ for moles of ${\text{Al}}^{3+}$ in the above equation as follows:

$\begin{array}{c}\text{number}{\text{of Al}}^{3+}\text{ions}=\left(4.1\times {10}^{-4}\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ Al}}^{3+}\text{ ions}}{1{\text{mole of Al}}^{3+}}\right)\\ =2.4680\times {10}^{20}\\ \approx 2.5\times {10}^{20}\end{array}$

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

$\text{amount}\text{of}{\text{SO}}_4^{2-}\left(\text{mol}\right)=\left({\text{moles of Al}}_2{\left({\text{SO}}_4\right)}_3\left(\text{mol}\right)\right)\left(\frac{{\text{moles of SO}}_4^{2-}\text{ion}\left(\text{mol}\right)}{1{\text{mole of Al}}_2{\left({\text{SO}}_4\right)}_3}\right)$

Substitute $2.06406\times {10}^{-4}\text{mol}$ for moles of ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3$ and $\text{3 mol}$ for moles of ${\text{SO}}_4^{2-}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{SO}}_4^{2-}\left(\text{mol}\right)=\left(2.06406\times {10}^{-4}\text{mol}\right)\left(\frac{\text{3 mol}}{1{\text{mole of Al}}_2{\left({\text{SO}}_4\right)}_3}\right)\\ =6.2\times {10}^{-4}\text{mol}\end{array}$

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 $6.2\times {10}^{-4}\text{mol}$ for moles of ${\text{SO}}_4^{2-}$ in the above equation as follows:

$\begin{array}{c}\text{number}{\text{of SO}}_4^{2-}\text{ions}=\left(6.2\times {10}^{-4}\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ SO}}_4^{2-}\text{ ions}}{1{\text{mole of SO}}_4^{2-}}\right)\\ =3.7289\times {10}^{20}\\ \approx 3.7\times {10}^{20}\end{array}$

Conclusion

The number of moles of is ${\text{Al}}^{3+}$ and ${\text{SO}}_4^{2-}$ is $4.1\times {10}^{-4}\text{mol}$ and $6.2\times {10}^{-4}\text{mol}$ respectively. The number of ions ${\text{Al}}^{3+}$ and ${\text{SO}}_4^{2-}$ is $2.5\times {10}^{20}$ and $3.7\times {10}^{20}$ respectively.

(c)

Interpretation Introduction

Interpretation:

The number of moles and number of ions of each type in $1.65\text{ mL}$ of a solution containing $8.83\times {10}^{21}$ formula units of cesium nitrate 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 4.27P

The number of moles of is ${\text{Cs}}^{+}$ and ${\text{NO}}_3^{-}$ is $\text{0}\text{.0242 mol}$ and $\text{0}\text{.0242 mol}$ respectively. The number of ions ${\text{Cs}}^{+}$ and ${\text{NO}}_3^{-}$ is $1.46\times {10}^{22}$ and $1.46\times {10}^{22}$ respectively.

Explanation of Solution

One mole of cesium nitrate $\left({\text{CsNO}}_3\right)$ on dissociation produces one mole of ${\text{Cs}}^{+}$ ion and one mole of ${\text{NO}}_3^{-}$ ion.

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

The expression to calculate the moles of ${\text{CsNO}}_3$ is:

${\text{moles of CsNO}}_3\left(\text{mol}\right)=\left[\begin{array}{c}\left(\text{volume of solution}\left(\text{mL}\right)\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ \left({\text{given formula unit of CsNO}}_3\left(\text{FU}\right)\right)\\ \left(\frac{{\text{1 mole of CsNO}}_3}{\text{6}\text{.022}\times {\text{10}}^{23}\text{FU}}\right)\end{array}\right]$

Substitute $8.83\times {10}^{21}$ formula unit for given formula unit of ${\text{CsNO}}_3$ and $1.65\text{ mL}$ for the volume of solution in the above equation as follows:

$\begin{array}{c}{\text{moles of CsNO}}_3\left(\text{mol}\right)=\left[\begin{array}{c}\left(1.65\text{ mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ \left(8.83\times {10}^{21}\right)\left(\frac{\text{1 mole of KBr}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{FU}}\right)\end{array}\right]\\ =0.024194\text{mol}\end{array}$

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

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

Substitute $0.024194\text{mol}$ for moles of ${\text{CsNO}}_3$ and $\text{1 mol}$ for moles of ${\text{Cs}}^{+}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{Cs}}^{+}\left(\text{mol}\right)=\left(0.024194\text{mol}\right)\left(\frac{\text{1 mol}}{1{\text{mole of CsNO}}_3}\right)\\ =0.0242\text{mol}\end{array}$

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

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

Substitute $0.0242\text{mol}$ for moles of ${\text{Cs}}^{+}$ in the above equation as follows:

$\begin{array}{c}\text{number}{\text{of Cs}}^{+}\text{ions}=\left(0.0242\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ Cs}}^{+}\text{ ions}}{1{\text{mole of Cs}}^{+}}\right)\\ =1.456\times {10}^{22}\\ \approx 1.46\times {10}^{22}\end{array}$

The expression to calculate the amount of ${\text{NO}}_3^{-}$ ion in moles is as follows:

$\text{amount}\text{of}{\text{NO}}_3^{-}\left(\text{mol}\right)=\left({\text{moles of CsNO}}_3\left(\text{mol}\right)\right)\left(\frac{{\text{moles of NO}}_3^{-}\text{ion}\left(\text{mol}\right)}{1{\text{mole of CsNO}}_3}\right)$

Substitute $0.024194\text{mol}$ for moles of ${\text{CsNO}}_3$ and $\text{1 mol}$ for moles of ${\text{NO}}_3^{-}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{NO}}_3^{-}\left(\text{mol}\right)=\left(0.024194\text{mol}\right)\left(\frac{\text{1 mol}}{1{\text{mole of CsNO}}_3}\right)\\ =0.0242\text{mol}\end{array}$

The expression to calculate the number of ${\text{NO}}_3^{-}$ ions is:

$\text{number}{\text{of NO}}_3^{-}\text{ions}=\left({\text{moles of NO}}_3^{-}\left(\text{mol}\right)\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{NO}}_3^{-}\text{ ions}}{1{\text{mole of NO}}_3^{-}}\right)$

Substitute $0.0242\text{mol}$ for moles of ${\text{NO}}_3^{-}$ in the above equation as follows:

$\begin{array}{c}\text{number}{\text{of NO}}_3^{-}\text{ions}=\left(0.0242\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ NO}}_3^{-}\text{ ions}}{1{\text{mole of NO}}_3^{-}}\right)\\ =1.4569\times {10}^{22}\\ \approx 1.46\times {10}^{22}\end{array}$

Conclusion

The number of moles of is ${\text{Cs}}^{+}$ and ${\text{NO}}_3^{-}$ is $\text{0}\text{.0242 mol}$ and $\text{0}\text{.0242 mol}$ respectively. The number of ions ${\text{Cs}}^{+}$ and ${\text{NO}}_3^{-}$ is $1.46\times {10}^{22}$ and $1.46\times {10}^{22}$ respectively.