Chapter 4, Problem 4.24P

(a)

Interpretation Introduction

Interpretation:

The mass $\left(\text{g}\right)$ of solute needed to make $475\text{ mL}$ of $5.62\times {10}^{-2}M$ potassium sulphate 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 amount of compound when moles of the compound and molecular mass are given:

$\text{amount of compound}\left(\text{g}\right)=\text{moles of compound}\left(\text{mol}\right)\left(\frac{\text{molecular mass of compound}\left(\text{g}\right)}{1\text{mole of compound}}\right)$

(b)

Interpretation Introduction

Interpretation:

Molarity of a solution that contains $7.25\text{ mg}$ of calcium chloride in each millilitre 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 molarity of a solution when moles of solute and volume of solution are given is as follows:

$\text{Molarity of solution}\left(M\right)=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$

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

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

(c)

Interpretation Introduction

Interpretation:

The number of ${\text{Mg}}^{2+}$ ions in each milliliters of $0.184M$ magnesium bromide 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 $\text{mol}/\text{L}$.

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 amount of ions in moles is as follows:

$\text{amount}\text{of}\text{ion}\left(\text{mol}\right)=\left(\text{moles of compound}\left(\text{mol}\right)\right)\left(\frac{\text{moles of ion}\left(\text{mol}\right)}{1\text{mole of compound}}\right)$

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

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