Question

At 300K , the vapor pressure of pure benzene${\mathbf{\text{(C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{6}}\mathbf{)}$ is$\mathbf{0}\mathbf{.}\mathbf{1355}\mathbf{}\mathbf{\text{atm}}$ and the vapor pressure of puren-hexane$\left({\text{C}}_6{\text{H}}_{14}\right)$ is$\mathbf{0}\mathbf{.}\mathbf{2128}$ atm. Mixing 50.0 g of benzene with 50.0 g of n-hexane gives a solution that is nearly ideal.

(a) Calculate the mole fraction of benzene in the solution.

(b) Calculate the total vapor pressure of the solution at 300 K .

(c) Calculate the mole fraction of benzene in the vapor in equilibrium with the solution.

Short answer

(a) The mole fraction of benzene in the solution is 52.4% .

(b) The total vapor pressure of the solution is 0.172 atm .

(c) The mole fraction of benzene in the vapor in equilibrium with the solution are 41.3% and 58.7%.

Step-by-step solution

Given data

The vapor pressure of benzene and n-hexane are 0.1355 atm and 0.2128 atm. The mass of both benzene and n-hexane is 50.0 g .

Concept of Raoult's law

According to Raoult's law, the partial vapor pressure of a solvent in a solution (or mixture) is equal to or the same as the vapor pressure of the pure solvent times the mole fraction of the solvent present in the solution.

The formula of Raoult's law is as follows:

${\mathit{p}}_{\mathbf{i}}\mathbf{=}{\mathit{X}}_{\mathbf{i}}\mathbf{\times }{\mathit{p}}_{\mathbf{i}}^{\mathbf{*}}$

Raoult's law to determine the mole fractions in the mixture as well as the total vapor pressure of it Finally, the composition of the vapor in equilibrium with the mixture is richer in the more volatile component The calculation uses the ratio of partial vapor pressures.

Calculate the mole fraction of benzene

(a)

The composition of the solution is equal masses (50 grams) of the two liquids mixed together. The number of moles is obtained by the molar mass formula as follows:

$$\begin{gathered} n=\frac{m}{M} \\ {\text{n(C}}_6{\text{H}}_6)=\frac{50.0}{78.0} \\ {\text{n(C}}_6{\text{H}}_6)=0.641 \end{gathered}$$

Similarly, number of moles of n-hexane is obtained as,

$$\begin{gathered} n_{{\text{C}}_6\text{H14}}=\frac{50.0}{86.0} \\ {n}_{{\text{C}}_6\text{H14}}=0.581\text{mol} \end{gathered}$$

from this, the mole fraction of benzene is:

$$\begin{gathered} X_{{\text{C}}_6{\text{H}}_6}=\frac{n_{{\text{C}}_8{\text{H}}_8}}{n} \\ {X}_{{\text{C}}_6{\text{H}}_6}=52.4\% \end{gathered}$$

Calculate total vapor pressure of the solution

(b)

The vapor pressures in ideal mixtures, is find by Raoult’s law in which vapor pressures is directly proportional to the mole fraction of the component.

The vapor pressures of the benzene is,

$$\begin{gathered} p_i=X_i\times p_i^{*}{p}_{{\text{C}}_8{\text{H}}_8} \\ =0.524\times 0.1355 \\ =0.071\text{atm} \end{gathered}$$

Similarly, the vapor pressures of the n-hexane is,

$$\begin{gathered} p_{{\text{C}}_6{\text{H}}_{14}}=(1-0.524)\times 0.2128 \\ =0.101\text{atm} \end{gathered}$$

Finally, By Dalton's law of partial pressures, the total vapor pressure is the sum of benzene and n-hexane.

$$\begin{gathered} p_v=0.071+0.101 \\ {p}_v=0.172\text{atm} \end{gathered}$$

Calculate the mole fraction of benzene in the vapor in equilibrium with the solution.

(c)

The mole fraction in vapor is obtained by the proportionality between partial pressures and the mole fraction, $Y_i=\frac{p_i}{p_v}$

$$\begin{gathered} Y_{{\text{C}}_6{\text{H}}_6}=\frac{0.071}{0.172} \\ {Y}_{{\text{C}}_6{\text{H}}_6}=41.3\% \\ {Y}_{{\text{C}}_8{\text{H}}_{14}}=\frac{0.101}{0.172}l \\ {Y}_{{\text{C}}_8{\text{H}}_{14}}=58.7\% \end{gathered}$$

It is clear that the ratio of the less volatile benzene dropped from $52.4\%$ (in liquid) to $41.3\%$ (in vapor).