Simple Distillation: What is the molar composition of the vapor in equilibrium with a boiling liquid that has a composition of 60% A and 40% B? I know that the vapor should be richer in the lower-boiling substance, which is A in this case.

And: If a sample of the vapor has the composition of 50% A and 50% B, what is the composition of the boiling liquid? Would the liquid composition also be 50% A and 50% B?
Can anyone explain?

To determine the molar composition of the vapor in equilibrium with a boiling liquid, you can use the concept of Raoult's Law. According to Raoult's Law, the partial pressure of a component in the vapor phase is directly proportional to its mole fraction in the liquid phase.

In the given scenario, the liquid has a composition of 60% A and 40% B. Since the vapor is richer in the lower-boiling substance (A), the mole fraction of A in the vapor will be higher than in the liquid. Let's calculate the molar composition of the vapor:

1. Convert the composition to mole fractions:
- Mole fraction of A in the liquid (X[A]_liquid) = 60% = 0.6
- Mole fraction of B in the liquid (X[B]_liquid) = 40% = 0.4

2. Apply Raoult's Law:
- Partial pressure of A in the vapor (P[A]_vapor) = X[A]_liquid * P[A]°
- Partial pressure of B in the vapor (P[B]_vapor) = X[B]_liquid * P[B]°

Here, P[A]° and P[B]° are the vapor pressure of pure components A and B, respectively.

Since the vapor should be richer in A, the mole fraction of A in the vapor (X[A]_vapor) will be higher than X[A]_liquid. Similarly, the mole fraction of B in the vapor (X[B]_vapor) will be lower than X[B]_liquid.

Now, let's consider the second part of the question. If the composition of the vapor is 50% A and 50% B, we can use the same approach to determine the composition of the boiling liquid.

1. Convert the composition of the vapor to mole fractions:
- Mole fraction of A in the vapor (X[A]_vapor) = 50% = 0.5
- Mole fraction of B in the vapor (X[B]_vapor) = 50% = 0.5

2. Apply Raoult's Law:
- Partial pressure of A in the vapor (P[A]_vapor) = X[A]_vapor * P[A]°
- Partial pressure of B in the vapor (P[B]_vapor) = X[B]_vapor * P[B]°

Since the composition of the vapor and liquid are connected by Raoult's Law, we can determine the composition of the boiling liquid by calculating the mole fractions of A and B in the liquid based on the partial pressures of the components in the vapor.

In conclusion, the composition of the boiling liquid will not be the same as the composition of the vapor. The composition of the liquid will depend on the partial pressures of the components in the vapor phase, which are governed by Raoult's Law.

In simple distillation, the molar composition of the vapor in equilibrium with a boiling liquid can be calculated using Raoult's law. According to Raoult's law, the partial pressure of each component in the vapor phase is directly proportional to its mole fraction in the liquid phase.

Let's consider the first question regarding the molar composition of the vapor. Given that the boiling liquid has a composition of 60% A and 40% B, we know that the vapor will be richer in the lower-boiling substance, which is A in this case.

Assuming ideal behavior, the mole fraction of component A in the vapor will be equal to its mole fraction in the liquid. Therefore, the vapor will have a composition of 60% A and 40% B.

Now, let's move on to the second question. If we have a sample of the vapor with a composition of 50% A and 50% B, we can determine the composition of the boiling liquid.

Since the vapor and liquid are in equilibrium, their partial pressures must be equal. According to Raoult's law, the partial pressure of each component is proportional to its mole fraction in the liquid phase.

Let's assume the molar composition of the boiling liquid is x% A and (100-x)% B. According to Raoult's law, the partial pressure of A in the vapor is its mole fraction in the liquid multiplied by its vapor pressure at that temperature. Similarly, the partial pressure of B in the vapor is its mole fraction in the liquid multiplied by its vapor pressure at that temperature.

If the vapor has a composition of 50% A and 50% B, it means that the partial pressures of A and B in the vapor are equal. Therefore, we can set up the following equation:

(x% A) * (vapor pressure of A) = [(100 - x)% A] * (vapor pressure of B)

Solving this equation will give us the composition of the boiling liquid, with x% A and (100-x)% B. So, in this case, the liquid composition will not be 50% A and 50% B. It will depend on the vapor pressures of A and B and their mole fractions in the vapor.

20