Solving a Mole Fraction Problem with X & Y @ 250°C

[MeKnos]

this problem should be simple, but i must be doing it wrong.

At 250 degrees Celsius, a certain liquid X has a vapor pressure of 900 mm Hg and a second liquid Y has a pressure of 600 mm Hg. What is the minimum mole fraction of X necessary for the mixture to boil at this temperature (assume atmosphere pressure is 1 atm).

MY WORK:

Since I know that the total pressure at this temperature is 1500 mm Hg, and only 760 mm Hg is needed, then only about 51% of the pressures are needed. But when I do (51%)(900) and (49%)(600), it doesn't add up to 760 mm Hg.

When I do guess and check, i get about 53.3% of compound X. How do you do this problem!
Thank you.

 

[Borek]

What is expression that describes total vapor pressure over the mixture?

Hint: total pressure is not 1500 mmHg.

 

[MeKnos]

What is expression that describes total vapor pressure over the mixture?

Hint: total pressure is not 1500 mmHg.

I think I got it, but I just want to confirm the answer.

Ptot= (Xa)Px + (1-Xa)Py.
760 = (Xa)900 + 600 - (Xa)600
760 - 600 = 160 = 300(Xa)

Xa = 160/300 = 53.33%?

 

[Borek]

Looks OK.