Chapter 16, Problem 83A

Interpretation Introduction

Interpretation: The mole fraction of ethanol is to be calculated when $25.0\text{g}$ of ethanol $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\right)$ and $40.0\text{g}$ of water is present.

Concept introduction: Mole fraction is a term that can be used to express solution concentration. The mole fraction of a particular solute is the ratio of the number of moles of solute present in the solution to the total number of moles present in the solution is the addition of the number of moles of solute and solvent.

The formula that can be used for the calculation of mole fraction is,

  $X_a=\frac{n_a}{n_a+n_b}$

Here, $n_a$ are the moles of solvent, $n_b$ are the moles of solute, and $X_a$ is the mole fraction of solute.

The expression for moles is as follows:

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

Answer to Problem 83A

The mole fraction of ethanol is $0.185$ .

Explanation of Solution

The mole fraction of ethanol can be calculated by knowing the number of moles of solute and solvent present in a solution.

The given solute is ethanol and its given mass is $25.0\text{g}$ . And the molecular mass is 46.

The given solvent is water and its mass is $40.0\text{g}$ . And the molecular mass is 18.

Substitute, $25.0\text{g}$ for the mass of $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\right)$ and $\text{ 46 g/mol}$ for molar mass of $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\right)$ in equation (1).

  $\begin{array}{c}\text{moles}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\right)=\frac{\text{ 25 g}}{46\text{ g/mol}}\\ =0.54\text{ mol}\end{array}$

Substitute, $40.0\text{g}$ for the mass of $\left({\text{H}}_2\text{O}\right)$ and $\text{ 18 g/mol}$ for molar mass of $\left({\text{H}}_2\text{O}\right)$ in equation (1).

  $\begin{array}{c}\text{moles}\left({\text{H}}_2\text{O}\right)=\frac{\text{ 40 g}}{\text{18 g/mol}}\\ =2.2\text{ mol}\end{array}$

Substitute $0.5$ for $n_a$ and $2.2$ for $n_b$ in the formula of mole fraction to calculate the mole fraction of ethanol, $X_a$ .

  $\begin{array}{c}{X}_a=\frac{0.5}{0.5+2.2\text{mol}}\\ =0.185\end{array}$

Conclusion

Therefore, the mole fraction of ethanol is $0.185$ .