At 25°C the vapour pressure of pure pentane is 511 torr and that of hexane is 150. torr.

What is the mole fraction of pentane in a pentane-hexane solution that has a vapour pressure of 333 torr at 25°C?

ppen = Xpen*Popentane

phex = Xhex*Pohexane

(511*Xpen) + (150*Xhex) = 333
(511*Xp) + 150(1-Xp) = 333
Solve for Xpentane and Xhexane.

Oh, mole fractions and vapor pressure! Let's break it down.

First, we need to figure out the mole fraction of pentane in the solution. To do that, we can use Raoult's law, which states that the vapor pressure of a solution is proportional to the mole fraction of each component.

The vapor pressure of the solution is given as 333 torr. The vapor pressure of pure pentane is 511 torr. So, the mole fraction of pentane can be calculated as follows:

Mole fraction of pentane = (Vapor pressure of pentane in solution) / (Vapor pressure of pure pentane)

Mole fraction of pentane = 333 torr / 511 torr

Mole fraction of pentane ≈ 0.651

So, the mole fraction of pentane in the pentane-hexane solution is approximately 0.651.

And remember, when it comes to humor, I'm always fractionally funny!

To determine the mole fraction of pentane in the solution, we can use Raoult's law, which states that the partial pressure of a component in a mixture is equal to the product of its mole fraction and the vapor pressure of the pure component.

Let's denote the mole fraction of pentane as x. The mole fraction of hexane would then be (1 - x), since the sum of the mole fractions equals 1.

According to Raoult's law, the vapor pressure of the pentane-hexane solution can be calculated using the following equation:

P_total = x * P_pentane + (1 - x) * P_hexane

We are given that the total vapor pressure of the solution is 333 torr, the vapor pressure of pure pentane is 511 torr, and the vapor pressure of pure hexane is 150 torr. Plugging these values into the equation, we get:

333 = x * 511 + (1 - x) * 150

Simplifying the equation:

333 = 511x + 150 - 150x

Combining like terms:

333 = 361x + 150

Subtracting 150 from both sides:

183 = 361x

Dividing both sides by 361:

x = 183/361

Calculating x:

x ≈ 0.507

Therefore, the mole fraction of pentane in the pentane-hexane solution is approximately 0.507.

To find the mole fraction of pentane in the solution, we can use Raoult's law, which states that the vapor pressure of a solution is the mole fraction of each component multiplied by its corresponding vapor pressure in its pure state.

First, let's assign the variables:
- P₁: vapor pressure of pentane in solution
- P₂: vapor pressure of hexane in solution
- X₁: mole fraction of pentane
- X₂: mole fraction of hexane

According to Raoult's law:
P₁ = X₁ * P₁₀
P₂ = X₂ * P₂₀

where P₁₀ and P₂₀ are the vapor pressures of pure pentane and hexane, respectively.

Given values:
P₁₀ (vapor pressure of pure pentane) = 511 torr
P₂₀ (vapor pressure of pure hexane) = 150 torr
P (vapor pressure of the solution) = 333 torr

We want to find X₁ (mole fraction of pentane).

Now, rearrange the equation for P₁ in terms of X₁:
P₁ = X₁ * P₁₀
X₁ = P₁ / P₁₀

Substitute the given values:
X₁ = 333 torr / 511 torr

Calculating X₁:
X₁ ≈ 0.651

Therefore, the mole fraction of pentane in the pentane-hexane solution is approximately 0.651.