Chapter 13, Problem 13.91P

(a)

Interpretation Introduction

Interpretation:

The moles of particles of ${\text{CuSO}}_4$ in $1\text{mL}$ of the solution is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $M$ and its unit is $\text{mol/L}$.

The formula to calculate the molarity of the solution is as follows:

$\text{Molarity}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$ (1)

Answer to Problem 13.91P

The moles of ${\text{CuSO}}_4$ in $1\text{mL}$ of the solution is $4\times {10}^{-5}\text{mol}$.

Explanation of Solution

Rearrange equation (1) to calculate the amount of solute as follows:

$\text{Amount}\text{of}\text{solute}=\left(\text{Molarity}\right)\left(\text{Volume of}\text{}\text{solution}\right)$ (2)

The dissociation of ${\text{CuSO}}_4$ occurs as follows:

${\text{CuSO}}_4\to {\text{Cu}}^{+}+{\text{SO}}_4^{2-}$

One mole of ${\text{CuSO}}_4$ to give one ${\text{Cu}}^{+}$ and one ${\text{SO}}_4^{2-}$ ion.

The formula to calculate the moles of particles of any compound is as follows:

$\begin{array}{c}\text{Moles}\text{of particles}=\left(\text{Molarity}\right)\left(\text{Volume of}\text{}\text{solution}\right)\\ \left(\frac{\text{moles}\text{of dissociated }\text{}\text{particles}}{\text{moles}\text{of}\text{}\text{compound}}\right)\end{array}$ (3)

Substitute $0.02M$ for the molarity, $1\text{mL}$ for the volume of solution, $\text{2 mol}$ for the moles of dissociated particles and $\text{1mol}$ for the moles of the compound in equation (3) to calculate the moles of particles of ${\text{CuSO}}_4$.

$\begin{array}{c}\text{Moles}{\text{of particles of CuSO}}_4=\left(\frac{0.02\text{}\text{}\text{mol}}{1\text{}\text{L}}\right)\left(\text{1}\text{mL}\right)\left(\frac{1\text{}\text{L}}{{10}^3\text{mL}}\right)\left(\frac{\text{2}\text{mol}}{\text{1}\text{}\text{}\text{ mol}}\right)\\ =4\times {10}^{-5}\text{mol}\end{array}$

Conclusion

The moles of ${\text{CuSO}}_4$ in $1\text{mL}$ of the solution is $4\times {10}^{-5}\text{mol}$.

(b)

Interpretation Introduction

Interpretation:

The moles of particles of $\text{Ba}{\left(\text{OH}\right)}_2$ in $1\text{mL}$ of the solution is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $M$ and its unit is $\text{mol/L}$.

The formula to calculate the molarity of the solution is as follows:

$\text{Molarity}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$ (1)

Answer to Problem 13.91P

The moles of particles of $\text{Ba}{\left(\text{OH}\right)}_2$ in $1\text{mL}$ of the solution is $\text{1}\times {10}^{-5}\text{mol}$.

Explanation of Solution

The dissociation of $\text{Ba}{\left(\text{OH}\right)}_2$ occurs as follows:

$\text{Ba}{\left(\text{OH}\right)}_2\to {\text{Ba}}^{2+}+2{\text{OH}}^{-}$

One mole of $\text{Ba}{\left(\text{OH}\right)}_2$ dissociates to give one ${\text{Ba}}^{2+}$ and two ${\text{OH}}^{-}$ ions.

Substitute $0.004M$ for the molarity, $1\text{mL}$ for the volume of solution, $\text{3}\text{mol}$ for the moles of dissociated particles and $\text{1mol}$ for the moles of the compound in equation (3) to calculate the moles of particles of $\text{Ba}{\left(\text{OH}\right)}_2$.

$\begin{array}{c}\text{Moles}\text{of particles of Ba}{\left(\text{OH}\right)}_2=\left(\frac{0.004\text{mol}}{1\text{}\text{L}}\right)\left(\text{1}\text{mL}\right)\left(\frac{1\text{}\text{L}}{{10}^3\text{mL}}\right)\left(\frac{\text{3}\text{mol}}{\text{1}\text{}\text{}\text{ mol}}\right)\\ =1.2\times {10}^{-5}\text{mol}\\ =\text{1}\times {10}^{-5}\text{mol}\end{array}$

Conclusion

The moles of particles of $\text{Ba}{\left(\text{OH}\right)}_2$ in $1\text{mL}$ of the solution is $\text{1}\times {10}^{-5}\text{mol}$.

(c)

Interpretation Introduction

Interpretation:

The moles of particles of ${\text{C}}_5{\text{H}}_5\text{N}$ in $1\text{mL}$ of the solution is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $M$ and its unit is $\text{mol/L}$.

The formula to calculate the molarity of the solution is as follows:

$\text{Molarity}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$ (1)

Answer to Problem 13.91P

The moles of particles of ${\text{C}}_5{\text{H}}_5\text{N}$ in $1\text{mL}$ of the solution is $8\times {10}^{-5}\text{mol}$.

Explanation of Solution

Pyridine is an aromatic compound so it does not dissociate into ions.

Substitute $0.08M$ for the molarity, $1\text{mL}$ for the volume of solution, $\text{1mol}$ for the moles of dissociated particles and $\text{1mol}$ for the moles of the compound in equation (3) to calculate the moles of particles of ${\text{C}}_5{\text{H}}_5\text{N}$.

$\begin{array}{c}\text{Moles}{\text{of particles of C}}_5{\text{H}}_5\text{N}=\left(\frac{0.08\text{mol}}{1\text{}\text{L}}\right)\left(\text{1}\text{mL}\right)\left(\frac{1\text{}\text{L}}{{10}^3\text{mL}}\right)\left(\frac{\text{1}\text{mol}}{\text{1}\text{}\text{}\text{ mol}}\right)\\ =8\times {10}^{-5}\text{mol}\end{array}$

Conclusion

The moles of particles of ${\text{C}}_5{\text{H}}_5\text{N}$ in $1\text{mL}$ of the solution is $8\times {10}^{-5}\text{mol}$.

(d)

Interpretation Introduction

Interpretation:

The moles of particles of ${\left({\text{NH}}_4\right)}_2{\text{CO}}_3$ in $1\text{mL}$ of the solution is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $M$ and its unit is $\text{mol/L}$.

The formula to calculate the molarity of the solution is as follows:

$\text{Molarity}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$ (1)

Answer to Problem 13.91P

The moles of particles of ${\left({\text{NH}}_4\right)}_2{\text{CO}}_3$ in $1\text{mL}$ of the solution is $\text{2}\times {10}^{-4}\text{mol}$.

Explanation of Solution

The dissociation of ${\left({\text{NH}}_4\right)}_2{\text{CO}}_3$ occurs as follows:

${\left({\text{NH}}_4\right)}_2{\text{CO}}_3\to 2{\text{NH}}_4^{+}+{\text{CO}}_3^{2-}$

One mole of ${\left({\text{NH}}_4\right)}_2{\text{CO}}_3$ dissociates to give one ${\text{CO}}_3^{2-}$ and two ${\text{NH}}_4^{+}$ ions.

Substitute $0.05M$ for the molarity, $1\text{mL}$ for the volume of solution, $\text{3}\text{mol}$ for the moles of dissociated particles and $\text{1mol}$ for the moles of the compound in equation (3) to calculate the moles of particles of ${\left({\text{NH}}_4\right)}_2{\text{CO}}_3$.

$\begin{array}{c}\text{Moles}\text{of particles of }{\left({\text{NH}}_4\right)}_2{\text{CO}}_3=\left(\frac{0.05\text{mol}}{1\text{}\text{L}}\right)\left(\text{1}\text{mL}\right)\left(\frac{1\text{}\text{L}}{{10}^3\text{mL}}\right)\left(\frac{\text{3}\text{mol}}{\text{1}\text{}\text{}\text{ mol}}\right)\\ =1.5\times {10}^{-4}\text{mol}\\ =\text{2}\times {10}^{-4}\text{mol}\end{array}$

Conclusion

The moles of particles of ${\left({\text{NH}}_4\right)}_2{\text{CO}}_3$ in $1\text{mL}$ of the solution is $\text{2}\times {10}^{-4}\text{mol}$.