Question #5cddd

1 Answer
Apr 18, 2016

Answer:

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^@"C"#, pure benzene has a vapor pressure of #"753 mmHg"# and pure toluene has a vapor pressure of #"290 mmHg"#.

#P_B^@ = "753 mmHg"#

#P_T^@ = "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_"total" = P_B + P_T" " " "color(orange)((1))#

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#

#color(blue)(|bar(ul(color(white)(a/a)"mole fraction" = chi = "mole percent"/ 100color(white)(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_"B gas" = 30/100 = 0.30#

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

#P_B = chi_"B gas" * P_"total"#

This will be equivalent to

#0.30 = P_B/P_"total"" " " "color(orange)((2))#

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.

#color(blue)(|bar(ul(color(white)(a/a)P_i = chi_"i mixture" xx P_i^@color(white)(a/a)|)))#

This means that you have

#P_B = chi_"B liquid" * P_B^@#

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

#chi_"T liquid" = 1 - chi_"B liquid"#

The partial pressure of toluene above the solution will be

#P_T = (1- chi_"B liquid") * P_T^@#

Use equation #color(orange)((1))# 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_"total" = chi_"B liquid" * P_B^@ + P_T^@ - chi_"B liquid" * P_T^@#

#color(purple)(|bar(ul(color(white)(a/a)color(black)(P_"total" = P_T^@ + (P_B^@ - P_T^@) * chi_"B liquid")color(white)(a/a)|)))#

Now use equation #color(orange)((2))# to write

#0.30 = (P_B)/(P_T^@ + (P_B^@ - P_T^@) * chi_"B liquid")#

But since #P_b = chi_"B liquid" * P_B^@#, you will have

#0.30 = (chi_"B liquid" * P_B^@)/(P_T^@ + (P_B^@ - P_T^@) * chi_"B liquid")#

Rearrange to solve for #chi_"B liquid"#

#0.30P_T^@ + 0.30 * (P_B^@ - P_T^@) * chi_"B liquid" = chi_"B liquid" * P_B^@#

This will get you

#chi_"B liquid" *( P_B^@ - 0.30P_B^@ + 0.30P_T^@) = 0.30P_T^@#

#color(green)(|bar(ul(color(white)(a/a)color(black)(chi_"B liquid" = (0.30P_T^@)/(0.70P_B^@ - 0.30P_T^@))color(white)(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_"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.