Chapter 16, Problem 16.1P

Interpretation Introduction

Interpretation:

The mole fraction of ethanol and water in a solution that is made up of $20.0g$ of ethanol and $80.0g$ of water has to be calculated.

Concept Introduction:

Mole fraction:

Mole fraction of a solute in a solution can be calculated as given below.

  ${\chi }_1=\frac{n_1}{n_{tot}}$

Where, $n_1$ is the number of moles of solute and $n_{tot}$ is the sum of moles of solute and solvent.

Answer to Problem 16.1P

The mole fraction of ethanol and water in the solution are $0.89and0.911$ respectively.

Explanation of Solution

Given that, a solution has been prepared by mixing of $20.0g$ of ethanol and $80.0g$ of water.  The molar mass of water and ethanol are $18.02g/moland46.07g/mol$ respectively.

The number of moles of water and ethanol can be calculated as given below.

  $\begin{array}{l}{n}_{water}=\frac{mass}{molarmass}=\frac{80.0g}{18.02g/mol}=4.44mol\\ \\ n_{ethanol}=\frac{mass}{molarmass}=\frac{20.0g}{46.07g/mol}=0.434mol\end{array}$

Now, the mole fraction of water and ethanol can be calculated as given below.

  $\begin{array}{l}{\chi }_{water}=\frac{n_{water}}{n_{water}+n_{ethanol}}=\frac{4.44mol}{4.44mol+0.434mol}=0.911\\ \\ {\chi }_{ethanol}=\frac{n_{ethanol}}{n_{water}+n_{ethanol}}=\frac{0.434mol}{4.44mol+0.434mol}=0.089\end{array}$

Therefore, the mole fraction of ethanol and water in the solution are $0.89and0.911$ respectively.