# Question 5cddd

Apr 18, 2016

Here's my take on this.

#### Explanation:

The idea here is that you need to use Raoult's Law and Dalton's Law of Partial Pressures to find a relationship between the mole fraction of benzene above the liquid and the mole fraction of benzene in the mixture.

So, you know that at ${80}^{\circ} \text{C}$, pure benzene has a vapor pressure of $\text{753 mmHg}$ and pure toluene has a vapor pressure of $\text{290 mmHg}$.

${P}_{B}^{\circ} = \text{753 mmHg}$

${P}_{T}^{\circ} = \text{290 mmHg}$

If you take ${P}_{B}$ to be the partial pressure of benzene above the liquid and ${P}_{T}$ to be the partial pressure of toluene above the liquid, you can use Dalton's Law of Partial Pressure to say that the total pressure above the mixture will be

${P}_{\text{total" = P_B + P_T" " " }} \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

As you know, the partial pressure of a gas that's part of a gaseous mixture can also be expressed using the gas' mole fraction, which is simply its mole percent divided by $100$

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \frac{\text{mole fraction" = chi = "mole percent}}{100} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you know that benzene has a 30% mole percent above the liquid, which means that its mole fraction will be

${\chi}_{\text{B gas}} = \frac{30}{100} = 0.30$

The partial pressure of benzene above the liquid can thus be written as

${P}_{B} = {\chi}_{\text{B gas" * P_"total}}$

This will be equivalent to

$0.30 = {P}_{B} / {P}_{\text{total"" " " }} \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

Now focus on finding the total pressure of the mixture by using the mole fractions of the two components in the liquid.

According to Raoult's Law, the partial pressure of a component $i$ of a liquid mixture above the solution can be written using the partial pressure of the pure component and the mole fraction it has in the mixture.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{i} = {\chi}_{\text{i mixture}} \times {P}_{i}^{\circ} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that you have

${P}_{B} = {\chi}_{\text{B liquid}} \cdot {P}_{B}^{\circ}$

Since the mixture only contains benzene and toluene, you can write the mole fraction of toluene as

${\chi}_{\text{T liquid" = 1 - chi_"B liquid}}$

The partial pressure of toluene above the solution will be

${P}_{T} = \left(1 - {\chi}_{\text{B liquid}}\right) \cdot {P}_{T}^{\circ}$

Use equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ to rewrite the total pressure above the liquid as

P_"total" = chi_"B liquid" * P_B^@ + (1-chi_"B lqiuid") * P_T^@

Rearrange to get

${P}_{\text{total" = chi_"B liquid" * P_B^@ + P_T^@ - chi_"B liquid}} \cdot {P}_{T}^{\circ}$

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{P}_{\text{total" = P_T^@ + (P_B^@ - P_T^@) * chi_"B liquid}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now use equation $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to write

$0.30 = \frac{{P}_{B}}{{P}_{T}^{\circ} + \left({P}_{B}^{\circ} - {P}_{T}^{\circ}\right) \cdot {\chi}_{\text{B liquid}}}$

But since ${P}_{b} = {\chi}_{\text{B liquid}} \cdot {P}_{B}^{\circ}$, you will have

$0.30 = \left({\chi}_{\text{B liquid" * P_B^@)/(P_T^@ + (P_B^@ - P_T^@) * chi_"B liquid}}\right)$

Rearrange to solve for ${\chi}_{\text{B liquid}}$

$0.30 {P}_{T}^{\circ} + 0.30 \cdot \left({P}_{B}^{\circ} - {P}_{T}^{\circ}\right) \cdot {\chi}_{\text{B liquid" = chi_"B liquid}} \cdot {P}_{B}^{\circ}$

This will get you

${\chi}_{\text{B liquid}} \cdot \left({P}_{B}^{\circ} - 0.30 {P}_{B}^{\circ} + 0.30 {P}_{T}^{\circ}\right) = 0.30 {P}_{T}^{\circ}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\chi}_{\text{B liquid}} = \frac{0.30 {P}_{T}^{\circ}}{0.70 {P}_{B}^{\circ} - 0.30 {P}_{T}^{\circ}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to get

chi_"B liquid" = (0.30 * 290 color(red)(cancel(color(black)("mmHg"))))/((0.70 * 753 - 0.30 * 290)color(red)(cancel(color(black)("mmHg")))) = 0.142

This means that the mole fraction of toluene will be

${\chi}_{\text{T liquid}} = 1 - 0.142 = 0.858$

Expressed using the mole percent of the two substances, your solution will contain

color(green)(|bar(ul(color(white)(a/a)"14% benzene"color(white)(a/a)|)))" " and " "color(green)(|bar(ul(color(white)(a/a)"86% toluene"color(white)(a/a)|)))#

I'll leave the answers rounded to two sig figs, despite the fact that you have one sig fig for the mole percent of benzene above the solution.