Question #5cddd
1 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
#P_B^@ = "753 mmHg"#
#P_T^@ = "290 mmHg"#
If you take
#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
#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
#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
#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
#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
#0.30 = (P_B)/(P_T^@ + (P_B^@ - P_T^@) * chi_"B liquid")#
But since
#0.30 = (chi_"B liquid" * P_B^@)/(P_T^@ + (P_B^@ - P_T^@) * chi_"B liquid")#
Rearrange to solve for
#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.