At 20. °C, the vapor pressure of toluene is 22 millimeters of mercury and that of benzene is 75 millimeters of mercury. An ideal solution, equimolar in toluene and benzene, is prepared. At 20. °C, what is the mole fraction of benzene in the vapor in equilibrium with this solution?

Well, I must say this question is truly leaving me gasping for breath! But fret not, I've got an answer that might just make you crack a smile.

To find the mole fraction of benzene in the vapor, we can use Raoult's law. This law states that the partial pressure of a component in a solution is proportional to its mole fraction. Simply put, it's like having a recipe where the amount of each ingredient is proportional to how much it contributes to the overall flavor.

So, using Raoult's law, we can calculate the mole fraction of benzene in the vapor by dividing the vapor pressure of benzene by the total vapor pressure of the solution.

Here comes the punchline:

If you're familiar with math, the formula for calculating mole fraction (X) is X = P(1) / (P(1) + P(2)). In this case, P(1) represents the vapor pressure of benzene (75 mmHg) and P(2) represents the vapor pressure of toluene (22 mmHg).

Calculating it out, we get X = 75 / (75 + 22), which simplifies to X ≈ 0.773.

In other words, the mole fraction of benzene in the vapor is approximately 0.773. So, it seems like benzene is stealing the show in this vapor!

I hope my answer brought a bit of humor to your day. If you have any more questions, don't hesitate to ask!

To find the mole fraction of benzene in the vapor in equilibrium with the solution, we can use Raoult's Law. According to Raoult's Law, the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution.

Step 1: Convert the vapor pressures to partial pressures:
The partial pressure of toluene in the vapor is 22 millimeters of mercury.
The partial pressure of benzene in the vapor is 75 millimeters of mercury.

Step 2: Calculate the mole fraction of benzene in the liquid phase:
The mole fraction of toluene in the liquid phase is 0.5 (assuming an equimolar solution of toluene and benzene).
The mole fraction of benzene in the liquid phase is also 0.5.

Step 3: Calculate the mole fraction of benzene in the vapor phase:
According to Raoult's Law, the mole fraction of benzene in the liquid phase is equal to its mole fraction in the vapor phase when the solution is in equilibrium.

Therefore, the mole fraction of benzene in the vapor in equilibrium with this solution is 0.5.

To find the mole fraction of benzene in the vapor in equilibrium with the solution, we can use Raoult's Law.

Raoult's Law states that the vapor pressure of a component in an ideal solution is equal to the product of its mole fraction in the solution and its pure component vapor pressure.

Let's denote the mole fraction of benzene as x₁ and the mole fraction of toluene as x₂.

According to Raoult's Law, the vapor pressure of benzene above the solution would be:

P₁ = x₁ * P₁°

Where P₁ is the vapor pressure of benzene in the solution and P₁° is the pure component vapor pressure of benzene.

Similarly, the vapor pressure of toluene above the solution would be:

P₂ = x₂ * P₂°

Where P₂ is the vapor pressure of toluene in the solution and P₂° is the pure component vapor pressure of toluene.

Since we are given the pure component vapor pressures (P₁° = 75 mmHg and P₂° = 22 mmHg) and we know that the solution is equimolar in toluene and benzene, we can assume that x₁ = x₂ = 0.5.

Using the above equations, we can solve for the mole fraction of benzene (x₁) in the vapor phase:

P₁ = x₁ * P₁°
x₁ = P₁ / P₁°

Plugging in the values, we get:

x₁ = 75 mmHg / (75 mmHg + 22 mmHg) = 75 mmHg / 97 mmHg

So, the mole fraction of benzene in the vapor in equilibrium with this solution is approximately 0.7732.

The mole fraction of each component = 0.5

The vapor pressure of the mixture is:
P = (0.5)(22) + (0.5)(75) = ____ mm
X(benzene vapor) = P(benz. vapor)/P
X(benzene vapor) = (0.5)(75)/P