Chapter 16.4, Problem 45LC

Interpretation Introduction

Interpretation: To evaluate the mole fraction of each component in the solution.

Concept introduction:

Mole fraction is defined as the number of moles of the specific component in solution over the total number of moles given in the solution.

  ${\text{X}}_{\text{A}}=\frac{\text{mol A}}{\text{mol A}+\text{mole B}}$ ;

  ${\text{X}}_{\text{B}}=\frac{\text{mol B}}{\text{mol A}+\text{mole B}}$

Answer to Problem 45LC

The mole fraction (X) of each component in the given solution of 2.50 mol of ethanoic acid in $10$

mol of water is:

  $\begin{array}{l}{X}_{solute}(C_2{H}_4{O}_2)=0.200\\ X_{solvent}(H_{2}O)=0.800\end{array}$

Explanation of Solution

Mole fraction is defined as the number of moles of the specific component in solution over the total number of moles given in the solution.

  ${\text{X}}_{\text{A}}=\frac{\text{mol A}}{\text{mol A}+\text{mole B}}$ ;

  ${\text{X}}_{\text{B}}=\frac{\text{mol B}}{\text{mol A}+\text{mole B}}$

Where,

• ${\text{X}}_{\text{A}}$ is the mole fraction of solute A.
• ${\text{X}}_{\text{B}}$ is the mole fraction of solute B.
• Mol A is the moles of solute A.
• Mol B is the moles of solute B.
• Mol A+ Mol B is the total number of moles of all the solvents.

  ${\text{X}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}}\text{=}\frac{{\text{moles of (C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}\text{) in solution}}{\text{total moles of solution}}$

  ${\text{X}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\frac{{\text{moles of (H}}_{\text{2}}\text{O) in solution}}{\text{total moles of solution}}$

Solute $=C_2{H}_4{O}_2$

Moles of solute $(C_2{H}_4{O}_2)=2.50mol$

Solvent $=H_{2}O$

Moles of solvent $(H_{2}O)=10mol$

Calculating the total moles of the solution:

  $\text{total moles of the solution = moles of solute + moles of solvent}$

Total moles of the solution $=2.50+10$

Total moles of the solution $=12.50mol$

By substituting the values to calculate the moles of mole fraction of each component:

Mole fraction of $C_2{H}_4{O}_2$

  $\begin{array}{l}{X}_{C_2{H}_4{O}_2}=\frac{2.50}{12.50}\\ X_{C_2{H}_4{O}_2}=0.200mol\end{array}$

Mole fraction of $H_{2}O$

  $\begin{array}{l}{X}_{H_{2}O}=\frac{10}{12.50}\\ X_{H_{2}O}=0.800mol\end{array}$

Conclusion

The mole fraction (X) of each component in the given solution of 2.50 mol of ethanoic acid in 10 mol of water is:

  $\begin{array}{l}{X}_{solute}(C_2{H}_4{O}_2)=0.200\\ X_{solvent}(H_{2}O)=0.800\end{array}$