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:

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

(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:

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

(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:

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