Question #655b9

2 Answers
May 16, 2016

Answer will be (2) 140 mm of Hg .

Explanation:

We have , A + B -------> AB

Mole fraction (X) of A = 0.8 ; X of B will be = 1 - 0.8 = 0.2

Now , we are given , V.P and of pure A and V.P of solution AB as 70 and 84 mm of Hg respectively .

As it forms an ideal solution , therefore ;

V.P of AB = Partial V.P of A + Partial V.P of B

Partial V.P of any component = X of that component * V.P of that component in pure state

Therefore , substituting the values in
V.P of AB = Partial V.P of A + Partial V.P of B ; we get ,

84 = 70 * 0.8 + V.P of pure B * 0.2

Solving this we get 140 mm of Hg as the V.P of pure B .

May 16, 2016

I get "140 mm Hg"140 mm Hg.


The temperature is just for context.

This question is basically asking you to use Raoult's law for ideal binary mixtures:

\mathbf(P_j = chi_jP_j^"*")

where:

  • P_j is the "non-pure" vapor pressure of the solution (i.e. when there is component i in it).
  • P_j^"*" is the vapor pressure of the pure component j (i.e. the solution with no component i).
  • chi_j is the \mathbf("mol") fraction of component j in solution. As chi_j -> 1, P_j -> P_j^"*".

You have:

  • P_A^"*" = "70 mm Hg"
  • chi_A = 0.8
  • P_"tot" = "84 mm Hg"
  • P_B^"*" = ?

in an ideal binary mixture of A and B. But this information is really all you need. Recall Dalton's law of partial pressures:

\mathbf(P_"tot" = P_A + P_B)

Now, combine this with Raoult's law (best for ideal binary mixtures!) to get:

P_"tot" = chi_AP_A^"*" + chi_BP_B^"*"

Here we can use the fact that chi_A + chi_B = 1 to get:

color(blue)(P_"tot") = chi_AP_A^"*" + (1 - chi_A)P_B^"*"

= chi_AP_A^"*" + P_B^"*" - chi_AP_B^"*"

Now you can solve for P_B^"*".

P_"tot" - chi_AP_A^"*" = P_B^"*"(1 - chi_A)

color(blue)(P_B^"*") = (P_"tot" - chi_AP_A^"*")/(1 - chi_A)

= (84 - 0.8*(70))/(1 - 0.8)

= (84 - 56)/(0.2)

= 28*5 = color(blue)("140 mm Hg")