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:

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

Answer to Problem 4.24P

The mass $\left(\text{g}\right)$ of solute needed to make $475\text{ mL}$ of $5.62\times {10}^{-2}M$ potassium sulphate is $4.65\text{g}$.

Explanation of Solution

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

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

The expression to calculate the moles of potassium sulphate $\left({\text{K}}_2{\text{SO}}_4\right)$ is:

${\text{Moles of K}}_2{\text{SO}}_4\left(\text{mol}\right)=\text{volume of solution}\left(\text{mL}\right)\left(\frac{{10}^{-3}\text{L}}{1\text{mL}}\right)\left(\frac{\text{molarity}\text{of}{\text{K}}_2{\text{SO}}_4\left(\text{mol}\right)}{\text{1L of solution}}\right)$

Substitute $475\text{ mL}$ for the volume of solution and $5.62\times {10}^{-2}\text{mol/L}$ for molarity of ${\text{K}}_2{\text{SO}}_4$ in the above equation as follows:

$\begin{array}{c}{\text{Moles of K}}_2{\text{SO}}_4\left(\text{mol}\right)=475\text{ mL}\left(\frac{{10}^{-3}\text{L}}{1\text{mL}}\right)\left(\frac{5.62\times {10}^{-2}\text{mol}}{\text{1L of solution}}\right)\\ =0.026695\text{mol}\end{array}$

The molecular mass of potassium sulphate $\left({\text{K}}_2{\text{SO}}_4\right)$ is $\text{174}\text{.26 g/mol}$.

The expression to calculate the amount of ${\text{K}}_2{\text{SO}}_4$ is:

${\text{amount of K}}_2{\text{SO}}_4\left(\text{g}\right)={\text{moles of K}}_2{\text{SO}}_4\left(\text{mol}\right)\left(\frac{{\text{molecular mass of K}}_2{\text{SO}}_4\left(\text{g}\right)}{1{\text{mole of K}}_2{\text{SO}}_4}\right)$

Substitute $0.026695\text{mol}$ for moles of ${\text{K}}_2{\text{SO}}_4$ and $\text{174}\text{.26 g/mol}$ for the molecular mass of ${\text{K}}_2{\text{SO}}_4$ in the above equation as follows:

$\begin{array}{c}{\text{amount of K}}_2{\text{SO}}_4\left(\text{g}\right)=0.026695\text{mol}\left(\frac{174.26\text{ g}}{1{\text{mole of K}}_2{\text{SO}}_4}\right)\\ =4.6519\text{g}\\ \approx 4.65\text{g}\end{array}$

Conclusion

The mass $\left(\text{g}\right)$ of solute needed to make $475\text{ mL}$ of $5.62\times {10}^{-2}M$ potassium sulphate is $4.65\text{g}$.

(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:

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

Answer to Problem 4.24P

Molarity of a solution that contains $7.25\text{ mg}$ of calcium chloride in each millilitre is $0.0653M$.

Explanation of Solution

The relation between $\text{mg}$ and $\text{g}$ is:

$\text{1}\text{g}=\text{1000 mg}$

The molecular mass of calcium chloride $\left({\text{CaCl}}_2\right)$ is $\text{110}\text{.98 g/mol}$.

The expression to calculate the moles of calcium chloride $\left({\text{CaCl}}_2\right)$ is:

${\text{Moles of CaCl}}_2\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}{\text{of CaCl}}_2\left(\text{mg}\right)\left(\frac{1\text{g}}{1000\text{mg}}\right)\\ \left(\frac{\text{1mole}{\text{of CaCl}}_2\left(\text{mol}\right)}{{\text{molecular mass of CaCl}}_2\left(\text{g}\right)}\right)\end{array}\right]$

Substitute $7.25\text{ mg}$ for given mass of ${\text{CaCl}}_2$ and $\text{110}\text{.98 g/mol}$ for the molecular mass of ${\text{CaCl}}_2$ in the above equation as follows:

$\begin{array}{c}{\text{Moles of CaCl}}_2\left(\text{mol}\right)=\left[7.25\text{ mg}\left(\frac{1\text{g}}{1000\text{mg}}\right)\left(\frac{\text{1mole}{\text{of CaCl}}_2\left(\text{mol}\right)}{110.98\text{ g}}\right)\right]\\ =6.5323\times {10}^{-5}\text{mol}\end{array}$

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

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

The expression to calculate the molarity of ${\text{CaCl}}_2$ is:

${\text{Molarity of CaCl}}_2\left(M\right)=\left(\frac{{\text{moles of CaCl}}_2\left(\text{mol}\right)}{{\text{volume of CaCl}}_2\left(\text{mL}\right)}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)$

Substitute $6.5323\times {10}^{-5}\text{mol}$ for moles of ${\text{CaCl}}_2$ and $1.0\text{ mL}$ for the volume of ${\text{CaCl}}_2$ and in the above equation as follows:

$\begin{array}{c}{\text{Molarity of CaCl}}_2\left(M\right)=\left(\frac{6.5323\times {10}^{-5}\text{mol}}{1.0\text{ mL}}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =0.06527M\\ \approx 0.0653M\end{array}$

Conclusion

Molarity of a solution that contains $7.25\text{ mg}$ of calcium chloride in each millilitre is $0.0653M$.

(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:

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

Answer to Problem 4.24P

The number of ${\text{Mg}}^{2+}$ ions in each milliliter of $0.184M$ magnesium bromide is $1.1\times {10}^{20}$.

Explanation of Solution

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

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

The expression to calculate the moles of magnesium bromide is:

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

Substitute $1\text{mL}$ for the volume of solution and $0.184\text{mol}/\text{L}$ for molarity of magnesium bromide in the above equation as follows:

$\begin{array}{c}\text{Moles of magnesium bromide}\left(\text{mol}\right)=1\text{mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(\frac{0.184\text{mol}}{\text{1L of solution}}\right)\\ =0.000184\text{mol}\end{array}$

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

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

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

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

Substitute $0.000184\text{mol}$ for moles of ${\text{MgBr}}_2$ and $\text{1 mol}$ for moles of ${\text{Mg}}^{2+}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{Mg}}^{2+}\left(\text{mol}\right)=\left(0.000184\text{mol}\right)\left(\frac{\text{1 mol}}{1{\text{mole of MgBr}}_2}\right)\\ =0.000184\text{mol}\end{array}$

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 $0.000184\text{mol}$ for moles of ${\text{Mg}}^{2+}$ in the above equation as follows:

$\begin{array}{c}\text{number}\text{of}{\text{Mg}}^{2+}\text{ions}=\left(0.000184\text{mol}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{Cu}}^{2+}\text{ions}}{1{\text{mole of Cu}}^{2+}}\right)\\ =1.1080\times {10}^{20}\\ \approx 1.11\times {10}^{20}\end{array}$

Conclusion

The number of ${\text{Mg}}^{2+}$ ions in each milliliter of $0.184M$ magnesium bromide is $1.1\times {10}^{20}$.