Chapter 13, Problem 59E

Interpretation Introduction

(a)

Interpretation:

The mass of solute in ${\text{FeCl}}_{\text{3}}$ solution having molarity $0.200M$ and volume $\text{2}\text{.25}\text{L}$ is to be calculated.

Concept introduction:

Molarity of a solution is the amount of solute dissolved in a $\text{1000}\text{mL}$ solution. It is also defined as the number of moles of solute per liter volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 59E

The mass of solute in ${\text{FeCl}}_{\text{3}}$ solution is $\text{0}\text{.866}\text{L}$.

Explanation of Solution

It is given that the volume of solution is $\text{2}\text{.25}\text{L}$ and molarity of ${\text{FeCl}}_{\text{3}}$ solution is $0.200M$.

The formula to determine molarity is shown below.

$M=\frac{\text{n}}{\text{V}}$ …(1)

Where

• $M$ is the molarity of a solution.

• $\text{n}$ is the number of moles of solute.

• $\text{V}$ is the volume of the solution.

The molar mass of iron is $55.85\text{g}{\text{mol}}^{-1}$.

The molar mass of chlorine is $35.45\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{FeCl}}_{\text{3}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{FeCl}}_{\text{3}}& =& \text{Molar}\text{mass}\text{of}\text{Fe}+\left(\text{3×Molar}\text{mass}\text{of}\text{Cl}\right)\\ & =& 55.85\text{g}{\text{mol}}^{-1}+\left(3\times 35.45\text{g}{\text{mol}}^{-1}\right)\\ & =& 55.85\text{g}{\text{mol}}^{-\text{1}}+106.35\text{g}{\text{mol}}^{-\text{1}}\\ & =& 162.2\text{g}{\text{mol}}^{-\text{1}}\end{array}$

The formula to calculate the number of moles is shown below.

$\text{n}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$ …(2)

The value of the given mass of solute is taken as $\text{X}$.

Substitute the value of the molar mass of solute as $162.2\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\text{n}=\frac{\text{X}}{162.2\text{g}{\text{mol}}^{-1}}$

Substitute the values of number of moles, molarity of solution as $0.200M$ and volume of the solution as $\text{2}\text{.25}\text{L}$ in equation (1) to calculate the mass of solute as shown below.

$\text{0}\text{.200}\text{M}=\frac{\text{X}}{\text{162}\text{.2}\text{g}{\text{mol}}^{-\text{1}}}\times \frac{\text{1}}{\text{2}\text{.25}\text{L}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}0.200\text{M}\times 162.2\text{g}{\text{mol}}^{-\text{1}}\times \text{2}\text{.25}\text{L}& =& \text{X}\\ \text{72}\text{.99}\text{g}& =& \text{X}\\ \text{X}& =& \text{72}\text{.99}\text{g}\\ & \approx & \text{73}\text{.0}\text{g}\end{array}$

Therefore, the mass of solute is $\text{73}\text{.0}\text{g}$.

Conclusion

The mass of solute in ${\text{FeCl}}_{\text{3}}$ solution is calculated as $\text{73}\text{.0}\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The mass of solute in ${\text{KIO}}_4$ solution having molarity $0.200M$ and volume $\text{2}\text{.25}\text{L}$ is to be calculated.

Concept introduction:

Molarity of a solution is the amount of solute dissolved in a $\text{1000}\text{mL}$ solution. It is also defined as the number of moles of solute per liter volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 59E

The mass of solute in ${\text{KIO}}_4$ solution is $\text{104}\text{g}$.

Explanation of Solution

It is given that the volume of solution is $\text{2}\text{.25}\text{L}$ and molarity of ${\text{KIO}}_4$ solution is $0.200M$.

The formula to determine molarity is shown below.

$M=\frac{\text{n}}{\text{V}}$ …(1)

Where

• $M$ is the molarity of a solution.

• $\text{n}$ is the number of moles of solute.

• $\text{V}$ is the volume of the solution.

The molar mass of potassium is $39.10\text{g}{\text{mol}}^{-1}$.

The molar mass of iodine is $126.90\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-\text{1}}$

Therefore, the molar mass of ${\text{KIO}}_4$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{KIO}}_4& =& \text{Molar}\text{mass}\text{of}\text{K}+\text{Molar}\text{mass}\text{of}\text{I}+\left(\text{4×Molar}\text{mass}\text{of}\text{O}\right)\\ & =& 39.10\text{g}{\text{mol}}^{-1}+126.90\text{g}{\text{mol}}^{-\text{1}}+\left(4\times 16.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 39.10\text{g}{\text{mol}}^{-\text{1}}+126.90\text{g}{\text{mol}}^{-\text{1}}+64.00\text{g}{\text{mol}}^{-\text{1}}\\ & =& 230\text{g}{\text{mol}}^{-\text{1}}\end{array}$

The formula to calculate the number of moles is shown below.

$\text{n}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$ …(2)

The value of the given mass of solute is taken as $\text{X}$

Substitute the value of the molar mass of solute as $230\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\text{n}=\frac{\text{X}}{230\text{g}{\text{mol}}^{-1}}$

Substitute the values of molarity of solution as $0.200M$ in equation and volume of the solution as $\text{2}\text{.25}\text{L}$ in equation (1) to calculate the mass of solute.

$\text{0}\text{.200}\text{M}=\frac{\text{X}}{230\text{g}{\text{mol}}^{-\text{1}}}\times \frac{\text{1}}{\text{2}\text{.25}\text{L}}$

Rearrange the above equation,

$\begin{array}{rll}0.200\text{M}\times 230\text{g}{\text{mol}}^{-\text{1}}\times \text{2}\text{.25}\text{L}& =& \text{X}\\ 103.5\text{g}& =& \text{X}\\ \text{X}& =& 103.5\text{g}\\ & \approx & 104\text{g}\end{array}$

Therefore, the mass of solute is $\text{104}\text{g}$.

Conclusion

The mass of solute in ${\text{KIO}}_4$ solution is $104\text{g}$.

Interpretation Introduction

(c)

Interpretation:

The mass of solute in ${\text{ZnSO}}_4$ solution having molarity $0.295M$ and volume $\text{50}\text{.0}\text{mL}$ is to be calculated.

Concept introduction:

Molarity of a solution is the amount of solute dissolved in a $\text{1000}\text{mL}$ solution. It is also defined as the number of moles of solute per liter volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 59E

The mass of solute in ${\text{ZnSO}}_4$ solution is $161.46\text{g}{\text{mol}}^{-1}$.

Explanation of Solution

It is given that the volume of solution is $\text{50}\text{.0}\text{mL}$ and molarity of ${\text{ZnSO}}_4$ solution is $0.295M$.

The relation between $\text{L}$ and $\text{mL}$ is given below.

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

The probable unit factors are given below.

$\frac{\text{1}\text{L}}{\text{1000}\text{mL}},\frac{\text{1000}\text{mL}}{\text{1}\text{L}}$

The unit factor to determine $\text{L}$ from $\text{mL}$ is given below.

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

Therefore, the volume in $\text{L}$ is calculated as shown below.

$\begin{array}{rll}\text{Volume}& =& 50.\text{0 mL}\times \frac{\text{1}\text{.0}\text{L}}{\text{1000}\text{mL}}\\ & =& 0.05\text{L}\end{array}$

The formula to determine molarity is shown below.

$M=\frac{\text{n}}{\text{V}}$ …(1)

Where

• $M$ is the molarity of a solution.

• $\text{n}$ is the number of moles of solute.

• $\text{V}$ is the volume of the solution.

The molar mass of zinc is $65.39\text{g}{\text{mol}}^{-1}$.

The molar mass of sulfur is $32.07\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-\text{1}}$

Therefore, the molar mass of ${\text{ZnSO}}_4$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{ZnSO}}_4& =& \text{Molar}\text{mass}\text{of}\text{Zn}+\text{Molar}\text{mass}\text{of}\text{S}+\left(\text{4×Molar}\text{mass}\text{of}\text{O}\right)\\ & =& 65.39\text{g}{\text{mol}}^{-1}+32.07\text{g}{\text{mol}}^{-\text{1}}+\left(4\times 16.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 65.39\text{g}{\text{mol}}^{-\text{1}}+32.07\text{g}{\text{mol}}^{-\text{1}}+64.00\text{g}{\text{mol}}^{-\text{1}}\\ & =& 161.46\text{g}{\text{mol}}^{-\text{1}}\end{array}$

The formula to calculate the number of moles is shown below.

$\text{n}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$ …(2)

The value of the given mass of solute is considered as $\text{X}$

Substitute the value of the molar mass of solute as $161.46\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\text{n}=\frac{\text{X}}{161.46\text{g}{\text{mol}}^{-1}}$

Substitute the value of volume of the solution as $\text{0}\text{.05}\text{L}$ and molarity of solution as $0.295M$ in equation (1) to calculate the mass of solute.

$\text{0}\text{.295}\text{M}=\frac{\text{X}}{161.46\text{g}{\text{mol}}^{-\text{1}}}\times \frac{\text{1}}{0.05\text{L}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}0.295\text{M}\times 161.46\text{g}{\text{mol}}^{\text{-1}}\times 0.05\text{L}& =& \text{X}\\ 2.38\text{g}& =& \text{X}\\ \text{X}& =& 2.38\text{g}\end{array}$

Therefore, the mass of solute is $2.38\text{g}$.

Conclusion

The mass of solute in ${\text{ZnSO}}_4$ solution is calculated as $2.38\text{g}$.

Interpretation Introduction

(d)

Interpretation:

The mass of solute in $\text{Ni}{\left({\text{NO}}_3\right)}_2$ solution having molarity $0.295M$ and volume $\text{50}\text{.0}\text{mL}$ is to be calculated.

Concept introduction:

Molarity of a solution is the amount of solute dissolved in a $\text{1000}\text{mL}$ solution. It is also defined as the number of moles of solute per liter volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 59E

The mass of solute in $\text{Ni}{\left({\text{NO}}_3\right)}_2$ solution is $2.69\text{g}$.

Explanation of Solution

It is given that the volume of solution is $\text{50}\text{.0}\text{mL}$ and molarity of $\text{Ni}{\left({\text{NO}}_3\right)}_2$ solution is $0.295M$.

The relation between $\text{L}$ and $\text{mL}$ is given below.

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

The probable unit factors are given below.

$\frac{\text{1}\text{L}}{\text{1000}\text{mL}},\frac{\text{1000}\text{mL}}{\text{1}\text{L}}$

The unit factor to determine $\text{L}$ from $\text{mL}$ is given below.

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

Therefore, the volume in $\text{L}$ is calculated as shown below.

$\begin{array}{rll}\text{Volume}& =& 50.\text{0 mL}\times \frac{\text{1}\text{.0}\text{L}}{\text{1000}\text{mL}}\\ & =& 0.05\text{L}\end{array}$

The formula to determine molarity is shown below.

$M=\frac{\text{n}}{\text{V}}$ …(1)

Where

• $M$ is the molarity of a solution.

• $\text{n}$ is the number of moles of solute.

• $\text{V}$ is the volume of the solution.

The molar mass of nickel is $58.69\text{g}{\text{mol}}^{-1}$.

The molar mass of nitrogen is $14.01\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-\text{1}}$

Therefore, the molar mass of $\text{Ni}{\left({\text{NO}}_3\right)}_2$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}\text{Ni}{\left({\text{NO}}_3\right)}_2& =& \text{Molar}\text{mass}\text{of}\text{Ni}+\left(2\times \left(\begin{array}{l}\text{Molar}\text{mass}\text{of}\text{N}+\\ 3\times \text{Molar}\text{mass}\text{of}\text{O}\end{array}\right)\right)\\ & =& 58.69\text{g}{\text{mol}}^{-1}+\left(2\times \left(14.01\text{g}{\text{mol}}^{\text{-1}}+\left(3\times 16.00\text{g}{\text{mol}}^{\text{-1}}\right)\right)\right)\\ & =& 58.69\text{g}{\text{mol}}^{-\text{1}}+\left(2\times \left(14.01\text{g}{\text{mol}}^{\text{-1}}+48.00\text{g}{\text{mol}}^{\text{-1}}\right)\right)\\ & =& 58.69\text{g}{\text{mol}}^{-\text{1}}+124.02\text{g}{\text{mol}}^{-\text{1}}\end{array}$

The above equation gives the molar mass of $\text{Ni}{\left({\text{NO}}_3\right)}_2$ to be $1\text{82}\text{.71}\text{g}{\text{mol}}^{-\text{1}}$.

The formula to calculate the number of moles is shown below.

$\text{n}=\frac{\text{Given}\text{mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$ …(2)

The value of the given mass of solute is taken as $\text{X}$

Substitute the value of the molar mass of solute as $\text{182}\text{.71}\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\text{n}=\frac{\text{X}}{182.71\text{g}{\text{mol}}^{-1}}$

Substitute the value of volume of the solution as $\text{0}\text{.05}\text{L}$ and molarity of solution as $0.295M$ in equation (1) to calculate the mass of the solute.

$\text{0}\text{.295}\text{M}=\frac{\text{X}}{182.71\text{g}{\text{mol}}^{-\text{1}}}\times \frac{\text{1}}{0.05\text{L}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}0.295\text{M}\times 182.71\text{g}{\text{mol}}^{-\text{1}}\times 0.05\text{L}& =& \text{X}\\ 2.69\text{g}& =& \text{X}\\ \text{X}& =& 2.69\text{g}\end{array}$

Therefore, the mass of solute is $2.69\text{g}$.

Conclusion

The mass of solute in $\text{Ni}{\left({\text{NO}}_3\right)}_2$ solution is calculated as $2.69\text{g}$.