# 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 thereiscomponent#i# in it).#P_j^"*"# is thevapor pressure of the pure component j(i.e. the solution withnocomponent#i# ).#chi_j# is the#\mathbf("mol")# fractionof 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 **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

#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")#