Chapter 4, Problem 4.19P

(a)

Interpretation Introduction

Interpretation:

Moles of ion released when $0.805\text{mol}$ of ${\text{Rb}}_2{\text{SO}}_4$ is dissolved in water are to be calculated.

Concept introduction:

A solution is a combination of two parts: solute and solvent. A solute is the substance that is present in small quantity and solvent is the substance in which solute is dissolved. When water acts as a solvent then it is known as an aqueous solution.

Ionic compounds are the compounds that are composed of charged ions. They are held together by electrostatic forces. These compounds dissociate into ions when dissolved in water. ${\text{Rb}}_2{\text{SO}}_4$ is an example of an ionic compound and dissociate into ions when dissolved in water. The dissociation reaction of ${\text{Rb}}_2{\text{SO}}_4$ is:

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

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

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

Answer to Problem 4.19P

$0.805\text{mol}$ of ${\text{Rb}}_2{\text{SO}}_4$ gives $2.42\text{mol}$ of ions in water.

Explanation of Solution

One mole of ${\text{Rb}}_2{\text{SO}}_4$ on dissociation produces two moles of ${\text{Rb}}^{+}$ ion and one mole of ${\text{SO}}_4^{2-}$ ion. Hence, total three moles of ions are produced.

The expression to calculate the moles of ions of ${\text{Rb}}_2{\text{SO}}_4$ is as follows:

${\text{moles of ions of Rb}}_2{\text{SO}}_4\left(\text{mol}\right)=\left({\text{moles of Rb}}_2{\text{SO}}_4\left(\text{mol}\right)\right)\left(\frac{\text{total moles of ions}\left(\text{mol}\right)}{1{\text{mole of Rb}}_2{\text{SO}}_4}\right)$

Substitute $0.805\text{mol}$ for moles of ${\text{Rb}}_2{\text{SO}}_4$ and $\text{3 mol}$ for total moles of ion in the above equation as follows:

$\begin{array}{c}{\text{moles of ions of Rb}}_2{\text{SO}}_4\left(\text{mol}\right)=\left(0.805\text{mol}\right)\left(\frac{\text{3 mol}}{1{\text{mole of Rb}}_2{\text{SO}}_4}\right)\\ =2.42\text{mol}\end{array}$

Conclusion

One mole of ${\text{Rb}}_2{\text{SO}}_4$ on dissociation produces two moles of ${\text{Rb}}^{+}$ ion and one mole of ${\text{SO}}_4^{2-}$ ion. $0.805\text{mol}$ of ${\text{Rb}}_2{\text{SO}}_4$ gives $2.42\text{mol}$ of ions in water.

(b)

Interpretation Introduction

Interpretation:

Moles of ion released when $3.85\times {10}^{-3}\text{g}$ of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is dissolved in water is to be calculated.

Concept introduction:

A solution is a combination of two parts: solute and solvent. A solute is the substance that is present in small quantity and solvent is the substance in which solute is dissolved. When water acts as a solvent then it is known as an aqueous solution.

Ionic compounds are the compounds that are composed of charged ions. They are held together by electrostatic forces. These compounds dissociate into ions when dissolved in water. $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is an example of an ionic compound and dissociate into ions when dissolved in water. The dissociation reaction of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is:

$\text{Ca}{\left({\text{NO}}_3\right)}_2\left(s\right)\to {\text{Ca}}^{2+}\left(aq\right)+2{\text{NO}}_3^{-}\left(aq\right)$

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

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

Answer to Problem 4.19P

$3.85\times {10}^{-3}\text{g}$ of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ gives $7.0\times {10}^{-5}\text{mol}$ of ions in water.

Explanation of Solution

One mole of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ on dissociation produces one mole of ${\text{Ca}}^{2+}$ ion and two moles of ${\text{NO}}_3^{-}$ ion. Hence, total three moles of ions are produced.

The molecular mass of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is $\text{164}\text{.10}\text{g/mol}$.

The expression to calculate the moles of ions in $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is as follows:

$\text{moles of ions in Ca}{\left({\text{NO}}_3\right)}_2\left(\text{mol}\right)=\left[\begin{array}{c}\left(\text{given mass of Ca}{\left({\text{NO}}_3\right)}_2\left(\text{g}\right)\right)\\ \left(\frac{\text{1 mole of Ca}{\left({\text{NO}}_3\right)}_2}{\text{molecular mass of Ca}{\left({\text{NO}}_3\right)}_2\left(\text{g}\right)}\right)\\ \left(\frac{\text{total moles of ion}\left(\text{mol}\right)}{1\text{mole of Ca}{\left({\text{NO}}_3\right)}_2}\right)\end{array}\right]$

Substitute $3.85\times {10}^{-3}\text{g}$ for given mass of $\text{Ca}{\left({\text{NO}}_3\right)}_2$, $\text{164}\text{.10}\text{g/mol}$ for the molecular mass of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ and $3\text{mol}$ for total moles of ion in the above equation as follows:

$\begin{array}{c}\text{moles of Ca}{\left({\text{NO}}_3\right)}_2\left(\text{mol}\right)=\left[\begin{array}{l}\left(3.85\times {10}^{-3}\text{g}\right)\left(\frac{\text{1 mole of Ca}{\left({\text{NO}}_3\right)}_2}{164.10\text{g}}\right)\\ \left(\frac{\text{3}\text{mol}}{1\text{mole of Ca}{\left({\text{NO}}_3\right)}_2}\right)\end{array}\right]\\ =7.038\times {10}^{-5}\text{mol}\\ \approx 7.0\times {10}^{-5}\text{mol}\end{array}$

Conclusion

One mole of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ on dissociation produces one mole of ${\text{Ca}}^{2+}$ ion and two moles of ${\text{NO}}_3^{-}$ ion. $3.85\times {10}^{-3}\text{g}$ of $\text{Ca}{\left({\text{NO}}_3\right)}_2$ gives $7.0\times {10}^{-5}\text{mol}$ of ions in water.

(c)

Interpretation Introduction

Interpretation:

Moles of ion released when $4.03\times {10}^{19}$ formula unit of $\text{Sr}{\left({\text{HCO}}_3\right)}_2$ is dissolved in water are to be calculated.

Concept introduction:

A solution is a combination of two parts: solute and solvent. A solute is the substance that is present in small quantity and solvent is the substance in which solute is dissolved. When water acts as a solvent then it is known as an aqueous solution.

Ionic compounds are the compounds that are composed of charged ions. They are held together by electrostatic forces. These compounds dissociate into ions when dissolved in water. $\text{Sr}{\left({\text{HCO}}_3\right)}_2$ is an example of an ionic compound and dissociate into ions when dissolved in water. The dissociation reaction of $\text{Sr}{\left({\text{HCO}}_3\right)}_2$ is:

$\text{Sr}{\left({\text{HCO}}_3\right)}_2\left(s\right)\to {\text{Sr}}^{2+}\left(aq\right)+2{\text{HCO}}_3^{-}\left(aq\right)$

A formula unit is used for the ionic compound to represent their empirical formula. The expression to calculate the moles of ions in a compound is as follows:

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

Answer to Problem 4.19P

$4.03\times {10}^{19}$ formula unit of $\text{Sr}{\left({\text{HCO}}_3\right)}_2$ gives $2.01\times {10}^{-4}\text{mol}$ of ions in water.

Explanation of Solution

One mole of $\text{Sr}{\left({\text{HCO}}_3\right)}_2$ on dissociation produces one mole of ${\text{Sr}}^{2+}$ ion and two moles of ${\text{HCO}}_3^{-}$ ion. Hence, total three moles of ions are produced.

The expression to calculate the moles of ions in $\text{Sr}{\left({\text{HCO}}_3\right)}_2$ is as follows:

$\text{moles of ions in Sr}{\left({\text{HCO}}_3\right)}_2\left(\text{mol}\right)=\left[\begin{array}{c}\left(\text{given formula unit of Sr}{\left({\text{HCO}}_3\right)}_2\left(\text{FU}\right)\right)\\ \left(\frac{\text{1 mole of Sr}{\left({\text{HCO}}_3\right)}_2}{\text{6}\text{.022}\times {\text{10}}^{23}\text{FU}}\right)\\ \left(\frac{\text{total moles of ion}\left(\text{mol}\right)}{1\text{mole of Sr}{\left({\text{HCO}}_3\right)}_2}\right)\end{array}\right]$

Substitute $4.03\times {10}^{19}$ formula unit for given formula unit of $\text{Sr}{\left({\text{HCO}}_3\right)}_2$ and $3\text{mol}$ for total moles of ion in the above equation as follows:

$\begin{array}{c}\text{moles of ions in Sr}{\left({\text{HCO}}_3\right)}_2\left(\text{mol}\right)=\left[\begin{array}{c}\left(4.03\times {10}^{19}\text{FU}\right)\\ \left(\frac{\text{1 mole of Sr}{\left({\text{HCO}}_3\right)}_2}{\text{6}\text{.022}\times {\text{10}}^{23}\text{FU}}\right)\left(\frac{\text{3 mol}}{1\text{mole of Sr}{\left({\text{HCO}}_3\right)}_2}\right)\end{array}\right]\\ =2.007\times {10}^{-4}\text{mol}\\ \approx 2.01\times {10}^{-4}\text{mol}\end{array}$

Conclusion

One mole of $\text{Sr}{\left({\text{HCO}}_3\right)}_2$ on dissociation produces one mole of ${\text{Sr}}^{2+}$ ion and two moles of ${\text{HCO}}_3^{-}$ ion and $4.03\times {10}^{19}$ formula unit of $\text{Sr}{\left({\text{HCO}}_3\right)}_2$ gives $2.01\times {10}^{-4}\text{mol}$ of ions in water.