Chapter 17, Problem 129CP

Interpretation Introduction

Interpretation: The mole fraction of benzene in the original solution needs to be determined.

Concept Introduction: The ratio of number of moles of a substance in a compound to the total number of moles of the compound is said to be mole fraction. The formula is:

  $\text{Mole fraction = }\frac{\text{Number of moles of a substance}}{\text{Total number of moles in compound}}$

Answer to Problem 129CP

The mole fraction of benzene in the original solution is $\text{0}\text{.286}$ .

Explanation of Solution

Given:

Vapor pressure of pure benzene is 750.0 torr and of pure toluene is 300.0 torr.

The mole fraction in the given vapor is 0.714.

Since, the sum of all mole fractions in a given mixture is 1 so, the mole fraction of toluene in solution II is:

  ${\text{X}}_{\text{benzene}}^{\text{II}}{\text{ + X}}_{\text{toluene}}^{\text{II}}\text{ = 1}$

Where ${\text{X}}_{\text{benzene}}^{\text{II}}$ is mole fraction of benzene in solution II and ${\text{X}}_{\text{toluene}}^{\text{II}}$ is mole fraction of toluene in solution II.

Substituting the value of mole fraction of benzene in solution II as:

  $\begin{array}{l}0.714{\text{ + X}}_{\text{toluene}}^{\text{II}}\text{ = 1}\\ {\text{X}}_{\text{toluene}}^{\text{II}}\text{ = 1 - }0.714\\ {\text{X}}_{\text{toluene}}^{\text{II}}\text{ = 0}\text{.286}\end{array}$

The formula for a partial pressure of a gas in mixture is:

  ${\text{P}}_{\text{A}}{\text{= X}}_{\text{A}}{\text{P}}_{\text{total}}$

Where ${\text{P}}_{\text{A}}$ is partial pressure of gas A, ${\text{X}}_{\text{A}}$ is mole fraction of gas A and ${\text{P}}_{\text{total}}$ is the total pressure of the mixture.

Rearranging the equation as:

  ${\text{X}}_{\text{A}}\text{ = }\frac{{\text{P}}_{\text{A}}}{{\text{P}}_{\text{total}}}$

The above equation is written in terms of benzene in solution II as:

  ${\text{X}}_{\text{benzene}}^{\text{II}}\text{ = }\frac{{\text{P}}_{\text{benzene}}^{\text{II}}}{{\text{P}}_{\text{benzene}}^{\text{II}}{\text{+P}}_{\text{toluene}}^{\text{II}}}$ - (1)

According to Raoult’s Law,

  ${\text{P}}_{\text{benzene}}^{\text{II}}{\text{= X}}_{\text{benzene}}{\text{P}}_{\text{benzene}}^{\text{o}}$

Where ${\text{P}}_{\text{benzene}}^{\text{o}}$ vapor pressure of pure benzene. So, the equation (1) is rewritten as:

  $\begin{array}{l}{\text{X}}_{\text{benzene}}^{\text{II}}\text{ = }\frac{{\text{X}}_{\text{benzene}}{\text{P}}_{\text{benzene}}^{\text{o}}}{{\text{X}}_{\text{benzene}}{\text{P}}_{\text{benzene}}^{\text{o}}{\text{+X}}_{\text{toluene}}{\text{P}}_{\text{toluene}}^{\text{o}}}\\ {\text{X}}_{\text{benzene}}^{\text{II}}\text{ = }\frac{{\text{X}}_{\text{benzene}}{\text{P}}_{\text{benzene}}^{\text{o}}}{{\text{X}}_{\text{benzene}}{\text{P}}_{\text{benzene}}^{\text{o}}{\text{+(1 - X}}_{\text{benzene}}{\text{)P}}_{\text{toluene}}^{\text{o}}}\left({\text{X}}_{\text{benzene}}+{\text{X}}_{\text{toluene}}=\text{ 1}\right)\end{array}$

The value for, ${\text{P}}_{\text{benzene}}^{\text{o}}\text{ = 750}\text{.0 torr}$ , ${\text{P}}_{\text{toluene}}^{\text{o}}\text{ = 300}\text{.0 torr}$ and ${\text{X}}_{\text{benzene}}^{\text{II}}\text{ = 0}\text{.714}$ . Substituting these values in above equation as:

  $\text{0}\text{.714 = }\frac{{\text{X}}_{\text{benzene}}\text{750}}{{\text{X}}_{\text{benzene}}{\text{750+(1 - X}}_{\text{benzene}}\text{)300}}$

Solving for ${\text{X}}_{\text{benzene}}$ as:

  $\begin{array}{l}\text{0}\text{.714 = }\frac{{\text{750X}}_{\text{benzene}}}{{\text{750X}}_{\text{benzene}}{\text{+ 300 - 300 X}}_{\text{benzene}}}\\ \text{0}\text{.714 = }\frac{{\text{750X}}_{\text{benzene}}}{{\text{450X}}_{\text{benzene}}\text{+ 300 }}\\ \text{0}\text{.714 = }\frac{{\text{150(5X}}_{\text{benzene}}\text{)}}{{\text{150(3X}}_{\text{benzene}}\text{+ 2) }}\\ \text{0}\text{.714 = }\frac{{\text{5X}}_{\text{benzene}}}{{\text{3X}}_{\text{benzene}}\text{+ 2 }}\\ \text{2}{\text{.142X}}_{\text{benzene}}\text{ + 1}{\text{.428 = 5X}}_{\text{benzene}}\\ \text{1}\text{.428 = 2}{\text{.858X}}_{\text{benzene}}\\ {\text{X}}_{\text{benzene}}\text{ = }\frac{\text{1}\text{.428}}{\text{2}\text{.858}}\\ {\text{X}}_{\text{benzene}}\text{ = 0}\text{.4997}\simeq \text{ 0}\text{.5}\end{array}$

Thus, the mole fraction of benzene in solution II is 0.5 and the mole fraction of toluene in solution II is (1.0 -0.5) is 0.5.

Now, in order to calculate the mole fraction of benzene in original solution:

Let ${\text{X}}_{\text{benzene}}^{\text{I}}$ is mole fraction of benzene in solution I and ${\text{X}}_{\text{toluene}}^{\text{I}}$ is mole fraction of toluene in solution I.

Now, the equation (1) for solution (I) can be written as:

  ${\text{X}}_{\text{benzene}}^{\text{I}}\text{ = }\frac{{\text{P}}_{\text{benzene}}^{\text{I}}}{{\text{P}}_{\text{benzene}}^{\text{I}}{\text{+ P}}_{\text{toluene}}^{\text{I}}}$

Rewriting the above equation using Raoult’s law as:

  ${\text{X}}_{\text{benzene}}^{\text{I}}\text{ = }\frac{{\text{X}}_{\text{benzene}}{\text{P}}_{\text{benzene}}^{\text{o}}}{{\text{X}}_{\text{benzene}}{\text{P}}_{\text{benzene}}^{\text{o}}{\text{+(1 - X}}_{\text{benzene}}{\text{)P}}_{\text{toluene}}^{\text{o}}}$

Since, the solution behaves ideally, so ${\text{X}}_{\text{benzene}}^{\text{I}}\text{ = 0}\text{.5}$ .

Substituting the values:

  $\text{0}\text{.5 = }\frac{{\text{X}}_{\text{benzene}}\text{750}}{{\text{X}}_{\text{benzene}}{\text{750+(1 - X}}_{\text{benzene}}\text{)300}}$

Solving for ${\text{X}}_{\text{benzene}}$ as:

  $\begin{array}{l}\text{0}\text{.5 = }\frac{{\text{750X}}_{\text{benzene}}}{{\text{750X}}_{\text{benzene}}{\text{+ 300 - 300 X}}_{\text{benzene}}}\\ \text{0}\text{.5 = }\frac{{\text{750X}}_{\text{benzene}}}{{\text{450X}}_{\text{benzene}}\text{+ 300 }}\\ \text{0}\mathrm{..}\text{5 = }\frac{{\text{150(5X}}_{\text{benzene}}\text{)}}{{\text{150(3X}}_{\text{benzene}}\text{+ 2) }}\\ \text{0}\text{.5 = }\frac{{\text{5X}}_{\text{benzene}}}{{\text{3X}}_{\text{benzene}}\text{+ 2 }}\\ \text{1}{\text{.5X}}_{\text{benzene}}{\text{ + 1 = 5X}}_{\text{benzene}}\\ \text{1 = 3}{\text{.5X}}_{\text{benzene}}\\ {\text{X}}_{\text{benzene}}\text{ = }\frac{\text{1}}{\text{3}\text{.5}}\\ {\text{X}}_{\text{benzene}}\text{ = 0}\text{.2857}\simeq \text{ 0}\text{.286}\end{array}$

Hence, the mole fraction of benzene in the original solution is $\text{0}\text{.286}$ .