Chapter 4, Problem 4F.1E

Interpretation Introduction

Interpretation:

The relative amounts of the two phases for a mixture with an overall mole fraction of phenol of 0.45 has to be determined using lever rule.

Concept introduction:

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture. Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$

The lever rule is mainly used to determine compositions of phases and the relative proportions of phases to each other in Binary diagrams of compounds and using the lever rule we can determine quantitatively the relative composition of a mixture in a two phase region in a phase diagram.

According to lever rule, we can write,

$n=n^{\text{'}}+n^{\text{'}\text{'}}\boxed{\begin{array}{l}Where,\\ n^{\text{'}}:Totalamountofmoleculeinonephase\\ n^{\text{'}\text{'}}:Totalamountintheotherphase\end{array}}$

The total amount of A in the sample is $n{x}_A$ where $x_A$ it is the overall mole fraction of A in the sample. The overall amount of A is also the sum of its amounts in the two phases, where it has the mole fractions $x_A^{\text{'}}and{x}_A^{\text{'}\text{'}}$ respectively.

Thus,

$\begin{array}{l}n{x}_A=n^{\text{'}}{x}_A^{\text{'}}+n^{\text{'}\text{'}}{x}_A^{\text{'}\text{'}}\\ n^{\text{'}}\left(x_A^{\text{'}}-x_A\right)=n^{\text{'}\text{'}}\left(x_A^{}-x_A^{\text{'}\text{'}}\right)\end{array}$