Question #655b9
2 Answers
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 .
I get
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 componenti in it).P_j^"*" is the vapor pressure of the pure component j (i.e. the solution with no componenti ).chi_j is the\mathbf("mol") fraction of componentj in solution. Aschi_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
\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
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_"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")