Chapter 17, Problem 17E

Interpretation Introduction

Interpretation: The mass percent, mole fraction, molality and molarity of toluene needs to be determined.

Concept introduction:

To calculate mass percent of component A of a solution having two components A and B, use the formula given below.

  $\text{Mass}\text{percent=}\frac{\text{Mass}\text{of}\text{A}}{\text{Mass}\text{of}\text{A}+\text{Mass}\text{of}\text{B}}\times 100\%$

To calculate mole fraction of component A of a solution having two components A and B use the formula given below.

  $\text{Mole}\text{fraction}\text{of A=}\frac{\text{Moles}\text{of}\text{A}}{\text{Moles}\text{of}\text{A}+\text{Moles}\text{of}\text{B}}$

To calculate moles use the below formula.

  $\text{Number}\text{of}\text{moles=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

To calculate molality, use the below formula.

  $\text{Molality=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Mass}\text{of}\text{solvent}\text{in}\text{kg}}$

To calculate molarity, use the below formula.

  $\text{Molarity=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{Litre}}$

Answer to Problem 17E

• Mass percent of toluene is $28.4\%$ .
• Mole fraction of toluene is 0.251.
• Molality of toluene is 4.31m
• Molarity of toluene is 2.691 M.

Explanation of Solution

The volume of toluene= $50\text{mL}$

The volume of benzene= $125\text{mL}$

Density of toluene= $0.867\text{g}{\text{cm}}^{\text{3}}$

Density of benzene= $0.874\text{g}{\text{cm}}^{\text{3}}$

Now calculate mass of tolune and benzene using the below formula.

  $\text{Density=}\frac{\text{Mass}}{\text{Volume}}$

For toluene,

  $\begin{array}{c}\text{Mass}=\text{Density×Volume}\\ \text{Mass}\text{of}\text{toluene}=0.867\text{g}{\text{cm}}^{\text{3}}\times \text{50}\text{mL}\\ \text{Mass}\text{of}\text{toluene}=43.35\text{g}\end{array}$

For benzene,

  $\begin{array}{c}\text{Mass}=\text{Density×Volume}\\ \text{Mass}\text{of}\text{benzene}=0.874\text{g}{\text{cm}}^{\text{3}}\times 12\text{5}\text{mL}\\ \text{Mass}\text{of}\text{benzene}=109.25\text{g}\end{array}$

  $\begin{array}{c}\text{Total}\text{mass=43}\text{.35}\text{g+109}\text{.25}\text{g}\\ \text{=152}\text{.6}\text{g}\end{array}$

  $\begin{array}{c}\text{Mass}\text{percent}\text{of tolune=}\frac{\text{Mass}\text{of}\text{tolune}}{\text{Mass}\text{of}\text{tolune}+\text{Mass}\text{of}\text{benzene}}\times 100\\ =\frac{43.35g}{43.35g+109.25\text{g}}\times 100\%\\ =28.4\%\end{array}$

So, mass percent of toluene is $28.4\%$.

To calculate mole fraction, moles of benzene and toluene should be calculated.

To calculate moles use the below formula.

  $\text{Number}\text{of}\text{moles=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$  ......(1)

For toluene,

Mass=43.35g

  $\begin{array}{c}\text{Molar mass of toluene}({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{3}})=7\times 12+8\times 1\\ =84+8\\ =92\text{g}{\text{mol}}^{\text{-1}}\end{array}$

Put the values in equation (1).

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{toluene=}\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{43.35\text{g}}{92\text{g}{\text{mol}}^{-1}}\\ =0.471\text{mol}\end{array}$

For benzene

Mass=109.25g

  $\begin{array}{c}\text{Molar mass of benzene}({\text{C}}_{\text{6}}{\text{H}}_6)=6\times 12+6\times 1\\ =72+6\\ =78\text{g}{\text{mol}}^{\text{-1}}\end{array}$

Put the values in equation (1).

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{benzene=}\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{109.25\text{g}}{78\text{g}{\text{mol}}^{-1}}\\ =1.40\text{mol}\end{array}$

Now calculate mole fraction of toluene,

  $\begin{array}{c}\text{Mole}\text{fraction}\text{of}\text{tolune=}\frac{\text{Moles}\text{of}\text{tolene}}{\text{Moles}\text{of}\text{tolene}+\text{Moles}\text{of}\text{benzene}}\\ =\frac{0.471}{0.471+1.40}\\ =\frac{0.471}{1.871}\\ =0.251\end{array}$

Hence, mole fraction of toluene is 0.251.

Now calculate molality of toluene.

Moles of toluene=0.471

Mass of benzene =109.25 g

  $\begin{array}{c}\text{Molality=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Mass}\text{of}\text{solvent}\text{in}\text{kg}}\\ \text{Molality}\text{of}\text{toluene=}\frac{0.471\text{mol}}{0.10925\text{kg}}\\ =4.31\text{m}\end{array}$

Hence, molality of toluene is 4.31m.

Now calculate molarity of toluene.

Moles of toluene=0.471

The volume of toluene= $50\text{mL}$

The volume of benzene= $125\text{mL}$

  $\begin{array}{c}\text{Total volume of solution}=50\text{mL+125}\text{mL}\\ \text{=175}\text{mL}\end{array}$

  $\begin{array}{c}\text{Molarity=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{Litre}}\\ =\frac{0.471\text{mol}}{0.175\text{L}}\\ =2.691\text{M}\end{array}$

Hence, molarity of toluene is 2.691 M.

Conclusion

• Mass percent of toluene is $28.4\%$ .
• Mole fraction of toluene is 0.251.
• Molality of toluene is 4.31m
• Molarity of toluene is 2.691 M.