Question #655b9

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 #"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")#