At 20 degrees celcius the vapor pressure of toluene is 22 mmhg and that of benzene is 75 mmhg. Assume an ideal solution, equimolar in benzene and toluene. At 20 degrees celcius what is the mole fraction of benzene in the vapor in equilibrium with this solution?

Question

At 20 degrees celcius the vapor pressure of toluene is 22 mmhg and that of benzene is 75 mmhg. Assume an ideal solution, equimolar in benzene and toluene. At 20 degrees celcius what is the mole fraction of benzene in the vapor in equilibrium with this solution?

If you use pv=nRt how would you find v and n. Or how do you solve this a different way

Answer 1

Note the correct spelling of celsius.

Equimolar means Xtol = Xben = 0.5

ptol = Xtol*Po toluene.
pben = Xben*P. benzene.
Ptotal = ptol + pben

Xtol vapor = ptol vapor/Ptotal
Xben vapor = pben vapor/Ptotal

Answer 2

To find the mole fraction of benzene in the vapor in equilibrium with the solution, you can use Raoult's law, which states that the vapor pressure of a component in an ideal solution is equal to the product of its mole fraction in the liquid phase and its pure vapor pressure.

First, let's define the variables:
Pbenzene = vapor pressure of benzene = 75 mmHg
Ptoluene = vapor pressure of toluene = 22 mmHg
Xbenzene = mole fraction of benzene in the liquid phase

According to Raoult's law, we have the equation:

Pbenzene = Xbenzene * Pbenzene_pure

Since the solution is equimolar in benzene and toluene, the mole fraction of benzene will be 0.5 (assuming equal amounts of benzene and toluene). Substituting the values into the equation:

75 mmHg = 0.5 * Pbenzene_pure

Now, solve for Pbenzene_pure:

Pbenzene_pure = 75 mmHg / 0.5
= 150 mmHg

So, at 20 degrees Celsius, the mole fraction of benzene in the vapor in equilibrium with the solution is 0.5.

Answer 3

To find the mole fraction of benzene in the vapor in equilibrium with the given solution, we can use Raoult's law, which states that the partial pressure of a component in a mixture is proportional to its mole fraction in the liquid phase.

First, let's calculate the mole fraction of benzene in the liquid phase. Since the solution is equimolar in benzene and toluene, the mole fraction of benzene would be 0.5, as it constitutes half of the total number of moles in the mixture.

Now, we can use Raoult's law to find the mole fraction of benzene in the vapor phase. The formula for Raoult's law is:

P_benzene = X_benzene * P_benzene^0

Where P_benzene is the partial pressure of benzene, X_benzene is its mole fraction in the liquid phase, and P_benzene^0 is the pure vapor pressure of benzene.

Rearranging the formula, we get:

X_benzene = P_benzene / P_benzene^0

Substituting the given values into the equation:

X_benzene = 75 mmHg / (75 mmHg + 22 mmHg)

Calculating this expression:

X_benzene = 75 mmHg / 97 mmHg

Simplifying:

X_benzene ≈ 0.773

Therefore, the mole fraction of benzene in the vapor in equilibrium with this solution at 20 degrees Celsius is approximately 0.773.