Vapour pressure in ideal solution?

Question

Calculate the mole fraction of liquid A in a mixture of two liquids, A and B, which has a total pressure of 29,0kPa, given that A and B form an ideal solution.
[Vapour pressure: A=15.0kPa, B=35.0kPa]

Answer 1

In an ideal solution the vapor pressure of the solution will be the sum of the vapor pressure of the componants, and the vapor pressure of the componants will be the vapor pressure of the pure componant times the mole fraction of that componant in the ideal solution.

In this solution, the partial pressure due to "A" will be the vapour pressure of pure A (15.0 kPa) times the mole fraction of A.

let x=mole fraction of A
since there are only two componants, then 1-x must equal the mole fraction of B

so:

x * 15.0 + (1-x) * 35.0 = 29.0
15x+35-35x=29
-20x=-6
x=6/20=3/10=.3

mole fraction of "A" is therefore .3

check:
.3*15= 4.5
.7*35=24.5
4.5+24.5=29

voila!

Answer 2

because of the fact the answer is right this is going to obey Raoult's regulation. So... enable x be the mole fraction of methanol. 174 = 303x + 40 4.6(a million-x) 129.4 = 258.4x So x = 0.50. So the mixture is an equimolar mix of methanol and propan-a million-ol