Chapter 13, Problem 60E

Interpretation Introduction

(a)

Interpretation:

The mass of solute in $\text{NaOH}$ solution having molarity $0.100M$ and volume $1.00\text{L}$ 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 litre volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 60E

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

Explanation of Solution

It is given that the volume of solution is $1.00\text{L}$ and molarity of $\text{NaOH}$ solution is $0.100M$.

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 sodium is $22.99\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $\text{1}\text{.01}\text{g}{\text{mol}}^{-\text{1}}$

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

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}\text{NaOH}& =& \text{Molar}\text{mass}\text{of}\text{Na}+\text{Molar}\text{mass}\text{of}\text{O}+\text{Molar}\text{mass}\text{of}\text{H}\\ & =& 22.99\text{g}{\text{mol}}^{-1}+16.00\text{g}{\text{mol}}^{-\text{1}}+1.01\text{g}{\text{mol}}^{-\text{1}}\\ & =& 40.0\text{g}{\text{mol}}^{-\text{1}}\end{array}$

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 $\text{40}\text{.0}\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\text{n}=\frac{\text{X}}{40.0\text{g}{\text{mol}}^{-1}}$

Substitute the values of number of moles, molarity of the solution as $0.200M$ and volume of the solution as $\text{0}\text{.025}\text{L}$ in equation (1) to calculate the mass of solute as shown below.

$\text{0}\text{.100}\text{M}=\frac{\text{X}}{40.0\text{g}{\text{mol}}^{-\text{1}}}\text{×}\frac{\text{1}}{1.00\text{L}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}0.100\text{M}\times 40.0\text{g}{\text{mol}}^{-\text{1}}\times 1.00\text{L}& =& \text{X}\\ 4.0\text{g}& =& \text{X}\\ \text{X}& =& 40.0\text{g}\end{array}$

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

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The mass of solute in ${\text{LiHCO}}_3$ solution having molarity $0.100M$ and volume $1.00\text{L}$ 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 litre volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 60E

The mass of solute in ${\text{LiHCO}}_3$ solution is $\text{6}\text{.796}\text{g}$.

Explanation of Solution

It is given that the volume of solution is $1.00\text{L}$ and molarity of $\text{NaOH}$ solution is $0.100M$.

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 lithium is $22.99\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $\text{1}\text{.01}\text{g}{\text{mol}}^{-\text{1}}$

The molar mass of carbon is $\text{12}\text{.01}\text{g}{\text{mol}}^{-\text{1}}$

Therefore, the molar mass of ${\text{LiHCO}}_3$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{LiHCO}}_3& =& \left(\left(\begin{array}{c}\text{Molar}\text{mass}\text{of}\text{Li}+\text{Molar}\text{mass}\text{of}\text{H}\\ +\text{Molar}\text{mass}\text{of}\text{C}+\left(\text{3}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\end{array}\right)\right)\\ & =& 6.94\text{g}{\text{mol}}^{-1}+1.01\text{g}{\text{mol}}^{-\text{1}}+12.01\text{g}{\text{mol}}^{-\text{1}}+\left(3\times 16.00\text{g}{\text{mol}}^{-\text{1}}\right)\\ & =& 6.94\text{g}{\text{mol}}^{-\text{1}}+1.01\text{g}{\text{mol}}^{-\text{1}}+12.01\text{g}{\text{mol}}^{-\text{1}}+48.00\text{g}{\text{mol}}^{-\text{1}}\\ & =& 67.96\text{g}{\text{mol}}^{-\text{1}}\end{array}$

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 $67.96\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\text{n}=\frac{\text{X}}{67.96\text{g}{\text{mol}}^{-1}}$

Substitute the values of number of moles and molarity of solution as $0.100M$ and volume of the solution as $\text{0}\text{.025}\text{L}$ in equation (1) to calculate the mass of solute as shown below.

$\text{0}\text{.100}\text{M}=\frac{\text{X}}{67.96\text{g}{\text{mol}}^{-\text{1}}}\text{×}\frac{\text{1}}{1.00\text{L}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}0.100\text{M}\times 67.96\text{g}{\text{mol}}^{-\text{1}}\times 1.00\text{L}& =& \text{X}\\ 6.796\text{g}& =& \text{X}\\ \text{X}& =& 6.796\text{g}\end{array}$

Therefore, the mass of solute is calculated as $6.796\text{g}$.

Conclusion

The mass of solute in ${\text{LiHCO}}_3$ solution is calculated as $6.796\text{g}$.

Interpretation Introduction

(c)

Interpretation:

The mass of solute in ${\text{CuCl}}_2$ solution having molarity $0.500M$ and volume $25.0\text{mL}$ 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 litre volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 60E

The mass of solute in ${\text{CuCl}}_2$ solution is $1.68\text{g}$.

Explanation of Solution

It is given that the volume of solution is $25.0\text{mL}$ and molarity of ${\text{CuCl}}_2$ solution is $0.500M$.

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.

$\begin{array}{rll}\text{Volume}& =& 25.\text{0 mL}\times \frac{\text{1}\text{.0}\text{L}}{\text{1000}\text{mL}}\\ & =& 0.025\text{L}\end{array}$

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 copper is $\text{63}\text{.55}\text{g}{\text{mol}}^{-1}$.

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

Therefore, the molar mass of ${\text{CuCl}}_2$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{CuCl}}_2& =& \text{Molar}\text{mass}\text{of}\text{Cu}+\left(\text{2}\times \text{Molar}\text{mass}\text{of}\text{Cl}\right)\\ & =& 63.55\text{g}{\text{mol}}^{-1}+\left(\text{2}\times 35.45\text{g}{\text{mol}}^{-\text{1}}\right)\\ & =& 63.55\text{g}{\text{mol}}^{-\text{1}}+70.9\text{g}{\text{mol}}^{-\text{1}}\\ & =& 134.45\text{g}{\text{mol}}^{-\text{1}}\end{array}$

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 $134.45\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\text{n}=\frac{\text{X}}{134.45\text{g}{\text{mol}}^{-1}}$

Substitute the values of number of moles, molarity of solution as $0.500M$ and volume of the solution as $\text{0}\text{.025}\text{L}$ in equation (1) to calculate the mass of solute as shown below.

$\text{0}\text{.500}\text{M}=\frac{\text{X}}{134.45\text{g}{\text{mol}}^{-1}}\text{×}\frac{\text{1}}{0.025\text{L}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}0.500\text{M}\times 134.45\text{g}{\text{mol}}^{-\text{1}}\times 0.025\text{L}& =& \text{X}\\ 1.68\text{g}& =& \text{X}\\ \text{X}& =& 1.68\text{g}\end{array}$

Therefore, the mass of solute is calculated as $1.68\text{g}$.

Conclusion

The mass of solute in ${\text{CuCl}}_2$ solution is calculated as $1.68\text{g}$.

Interpretation Introduction

(d)

Interpretation:

The mass of solute in ${\text{KMnO}}_4$ solution having molarity $0.500M$ and volume $25.0\text{mL}$ 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 litre volume of solution. Molarity is denoted by $M$ and it is independent of temperature.

Answer to Problem 60E

The mass of solute in ${\text{KMnO}}_4$ solution is $1.98\text{g}$.

Explanation of Solution

It is given that the volume of solution is $25.0\text{mL}$ and molarity of ${\text{KMnO}}_4$ solution is $0.500M$.

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.

$\begin{array}{rll}\text{Volume}& =& 25.\text{0 mL}\times \frac{\text{1}\text{.0}\text{L}}{\text{1000}\text{mL}}\\ & =& 0.025\text{L}\end{array}$

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 potassium is $39.10\text{g}{\text{mol}}^{-1}$.

The molar mass of manganese is $\text{54}\text{.94}\text{g}{\text{mol}}^{-\text{1}}$

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-\text{1}}$

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

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{KMnO}}_4& =& \text{Molar}\text{mass}\text{of}\text{K}+\text{Molar}\text{mass}\text{of}\text{Mn}+\left(\text{4}\times \text{Molar}\text{mass}\text{of}\text{O}\right)\\ & =& 39.10\text{g}{\text{mol}}^{-1}+54.94\text{g}{\text{mol}}^{-\text{1}}+\left(4\times 16.00\text{g}{\text{mol}}^{-\text{1}}\right)\\ & =& 39.10\text{g}{\text{mol}}^{-\text{1}}+54.94\text{g}{\text{mol}}^{-\text{1}}+64.00\text{g}{\text{mol}}^{-\text{1}}\\ & =& 158.04\text{g}{\text{mol}}^{-\text{1}}\end{array}$

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 $158.04\text{g}{\text{mol}}^{-1}$ in the equation (2).

$\text{n}=\frac{\text{X}}{158.04\text{g}{\text{mol}}^{-1}}$

Substitute the values of number of moles, molarity of the solution as $0.500M$ and volume of the solution as $\text{0}\text{.025}\text{L}$ in equation (1) to calculate the mass of solute as shown below.

$\text{0}\text{.500}\text{M}=\frac{\text{X}}{158.04\text{g}{\text{mol}}^{-1}}\times \frac{\text{1}}{0.025\text{L}}$

Rearrange the above equation as shown below.

$\begin{array}{rll}0.500\text{M}\times 158.04\text{g}{\text{mol}}^{-\text{1}}\times 0.025\text{L}& =& \text{X}\\ 1.975\text{g}& =& \text{X}\\ \text{X}& =& 1.975\text{g}\\ & \approx & \text{1}\text{.98}\text{g}\end{array}$

Therefore, the mass of solute is calculated as $1.98\text{g}$.

Conclusion

The mass of solute in ${\text{KMnO}}_4$ solution is calculated as $1.98\text{g}$.