At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene (vapor pressure of pure toluene = 290 torr). The mole fraction of benzene in the vapor above the solution is 0.590. Assuming ideal behavior, calculate the mole fraction of toluene in the solution.

Answer

احا

0.213

كل ما سبق لاغ

To calculate the mole fraction of toluene in the solution, we can use Raoult's law, which states that the vapor pressure of each component in a mixture is directly proportional to its mole fraction in the solution.

According to Raoult's law, the mole fraction of benzene in the vapor above the solution (Xbenzene) is 0.590. Since the mole fraction of toluene in the solution (Xtoluene) is equal to 1 - Xbenzene, we can substitute the given value into the equation to calculate Xtoluene.

Xtoluene = 1 - Xbenzene
Xtoluene = 1 - 0.590
Xtoluene = 0.410

Therefore, the mole fraction of toluene in the solution is 0.410.

PA=XAP'A

Since the benzene comes out better you will need more Toluene in the solution

to get its mol fraction above the solution
.59 = Pb/[(Pb)(Pt)}
Assumming 1 atm external pressure

.59 = (1-x)745/(1-x)745 + x(290)
=0.641