Chapter 4, Problem 4.22P

(a)

Interpretation Introduction

Interpretation:

The overall mole fraction of aniline in the given mixture has to be determined.

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,

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

Explanation of Solution

Given,

Aniline and hexane form partially miscible liquid-liquid mixtures at temperatures below $69.1{}^{o}C$. When $42.8g$ of aniline and $75.2g$ of hexane are mixed together at a temperature of ${67.5}^{o}C$, two separate liquid phases are formed, with mole fractions of aniline 0.308 and 0.618.

$\begin{array}{c}\text{The number of moles of aniline}\text{in}\text{the}\text{mixture}=\frac{\text{Given}\text{mass}\text{of}\text{aniline}\text{}}{\text{Molar}\text{mass}\text{of}\text{aniline}}\\ =\frac{\text{42}\text{.8}\text{g}}{\text{93}\text{.13}\text{g/mol}}\\ =0.460mol\\ \text{The number of moles of hexane}\text{in}\text{the}\text{mixture}=\frac{\text{Given}\text{mass}\text{of}\text{hexane}\text{}}{\text{Molar}\text{mass}\text{ of}\text{hexane}}\\ =\frac{75.2\text{g}}{86.18\text{g/mol}}\\ =0.873mol\end{array}$

Now, the mole fractions of aniline in this mixture can be determined,

$\begin{array}{c}Molefractionofaniline\left(x_A\right)=\frac{Molesofanilineinthemixture}{Totalmolesinthemixture}\\ =\frac{0.460}{0.460+0.873}\\ =0.345\end{array}$

(b)

Interpretation Introduction

Interpretation:

The relative amounts of two phases in the given mixture 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,

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

Explanation of Solution

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.

Given,

Aniline and hexane form partially miscible liquid-liquid mixtures at temperatures below $69.1{}^{o}C$. When $42.8g$ of aniline and $75.2g$ of hexane are mixed together at a temperature of ${67.5}^{o}C$, two separate liquid phases are formed, with mole fractions of aniline 0.308 and 0.618.

According to lever rule,

$n^{\text{'}}\left(x_A^{\text{'}}-x_A\right)=n^{\text{'}\text{'}}\left(x_A^{}-x_A^{\text{'}\text{'}}\right)$

Substituting these values in the above equation,

$\begin{array}{c}{n}^{\text{'}}\left(0.618-0.345\right)=n^{\text{'}\text{'}}\left(0.345-0.308\right)\\ n^{\text{'}}\left(0.273\right)=n^{\text{'}\text{'}}\left(0.037\right)\\ \frac{n^{\text{'}}}{n^{\text{'}\text{'}}}=\left(\frac{0.037}{0.273}\right)\\ \\ =0.135\end{array}$