Chapter SM, Problem 6PQ

Interpretation Introduction

Interpretation:

The vapor pressure of solution with the benzene to octane molar ratio of $2:1$ has to be determined.

Concept Introduction:

According to Raoult’s law, partial pressure of each component of an ideal mixture is determined by multiplication of vapor pressure of its pure component and its mole fraction in solution.

Mathematical expression for Raoult’s law is as follows:

  $P_{\text{solvent}}=X_{\text{solvent}}{P}_{\text{solvent}}^{°}$        (1)

Here,

  $P_{\text{solvent}}$ is vapor pressure of solvent.

  $X_{\text{solvent}}$ is mole fraction of solvent.

  $P_{\text{solvent}}^{°}$ is vapor pressure of pure solvent.

Raoult’s law also states that total vapor pressure of solution is calculated by sum of partial pressures of both components.

The expression for total pressure of solution is as follows:

  $P=P_{\text{A}}+P_{\text{B}}$        (2)

Here,

  $P$ is total pressure of solution.

  $P_{\text{A}}$ is partial pressure of component A.

  $P_{\text{B}}$ is partial pressure of component B.

Answer to Problem 6PQ

Option (B) is the correct one.

Explanation of Solution

Reason for correct option:

Since molar ratio of benzene to octane is $2:1$, number of moles of benzene and octane is considered to be $2x$ and $x$ respectively.

Total number of moles can be calculated as follows:

  $\text{Total number of moles}=\text{Moles of benzene}+\text{Moles of octane}$        (3)

Substitute $2x$ for moles of benzene and $x$ for moles of octane in equation (3).

  $\begin{array}{c}\text{Total number of moles}=\text{2}x+x\\ =3x\end{array}$

The formula to calculate mole fraction of component is as follows:

  $\text{Mole fraction}\left(X\right)=\frac{\text{Number of moles of component}}{\text{Total number of moles}}$        (4)

Substitute $2x$ for number of moles of component and $3x$ for total number of moles in equation (4) to calculate mole fraction of benzene.

  $\begin{array}{c}\text{Mole fraction of benzene}=\frac{\text{2}x}{\text{3}x}\\ =0.667\end{array}$

Substitute $x$ for number of moles of component and $3x$ for total number of moles in equation (4) to calculate mole fraction of octane.

  $\begin{array}{c}\text{Mole fraction of octane}=\frac{x}{\text{3}x}\\ =0.333\end{array}$

Substitute 0.667 for $X_{\text{solvent}}$ and $280\text{ mm Hg}$ for $P_{\text{solvent}}^{°}$ in equation (1) to calculate vapor pressure of benzene.

  $\begin{array}{c}{P}_{\text{benzene}}=\left(0.667\right)\left(280\text{ mm Hg}\right)\\ =186.76\text{ mm Hg}\end{array}$

Substitute 0.333 for $X_{\text{solvent}}$ and $400\text{ mm Hg}$ for $P_{\text{solvent}}^{°}$ in equation (1) to calculate vapor pressure of octane.

  $\begin{array}{c}{P}_{\text{octane}}=\left(0.333\right)\left(400\text{ mm Hg}\right)\\ =133.2\text{ mm Hg}\end{array}$

Given solution is mixture of benzene and octane. Therefore equation (2) modifies as follows:

  $P=P_{\text{benzene}}+P_{\text{octane}}$        (5)

Here,

  $P_{\text{benzene}}$ is vapor pressure of benzene.

  $P_{\text{octane}}$ is vapor pressure of octane.

Substitute $186.76\text{ mm Hg}$ for $P_{\text{benzene}}$ and $133.2\text{ mm Hg}$ for $P_{\text{octane}}$ in equation (5).

  $\begin{array}{c}P=186.76\text{ mm Hg}+133.2\text{ mm Hg}\\ =319.96\text{ mm Hg}\\ \approx \text{320 mm Hg}\end{array}$

Hence option (B) is correct.

Reason for incorrect option:

Vapor pressure of given solution comes out to be $\text{320 mm Hg}$ that does not match with $\text{120 mm Hg}$, $\text{400 mm Hg}$ and $\text{680 mm Hg}$. Hence option (A), (C) and (D) are incorrect.

Conclusion

The vapor pressure of solution with benzene to octane molar ratio of $2:1$ is $\text{320 mm Hg}$. Hence, correct option is (B).