Question

You have a solution of two volatile liquids, \(A\) and \(B\) (assume ideal behavior). Pure liquid A has a vapor pressure of 350.0 torr and pure liquid B has a vapor pressure of 100.0 torr at the temperature of the solution. The vapor at equilibrium above the solution has double the mole fraction of substance A that the solution does. What is the mole fraction of liquid A in the solution?

Short answer

The mole fraction of liquid A in the solution is \(\frac{1}{3}\), or approximately 0.333.

Step-by-step solution

Write Raoult's law for both liquids A and B

Raoult's law for liquid A can be written as: \(P_A = x_A \times P_A^\star\) And for liquid B, it can be written as: \(P_B = x_B \times P_B^\star\) Here, \(P_A\) and \(P_B\) are the partial pressures of liquids A and B in the vapor phase, whereas \(P_A^\star\) and \(P_B^\star\) are their vapor pressures when they are pure liquids. In our case, \(P_A^\star = 350.0\,\text{torr}\) and \(P_B^\star = 100.0\,\text{torr}\).

Calculate mole fraction of A in vapor phase

It is given that the mole fraction of A in the vapor phase is double that in the solution. Thus, we have: \(y_A = 2x_A\)

Calculate mole fraction of B in the vapor phase

Since the sum of mole fractions in the vapor phase must be equal to 1, we can write: \(y_B = 1 - y_A\)

Write the expressions for partial pressures of A and B using their mole fractions

Using the expressions for mole fractions in step 2 and step 3, we can write the expressions for partial pressures of A and B as: \(P_A = (2x_A) P_{total}\) \(P_B = (1 - 2x_A) P_{total}\)

Apply Raoult's law for A and B

Applying Raoult's law, we obtain the following equations: \(P_A = x_A \times 350.0\) \(P_B = x_B \times 100.0\) Since \(x_A\) + \(x_B\) = \(1\), we can write the Raoult's law equation for B as: \(P_B = (1 - x_A) \times 100.0\)

Solve for the mole fraction of A in the solution

Substitute the equations from steps 4 and 5 into each other to find the mole fraction of A in the solution: \(x_A \times 350.0 = (2x_A) P_{total}\) \((1 - x_A) \times 100.0 = (1 - 2x_A) P_{total}\) Divide the first equation by 350 and the second equation by -100: \(x_A = \frac{2}{350} P_{total}\) \(x_A - 1 = \frac{1}{100} P_{total}\) Now we can combine the two equations: \(\frac{2}{350} P_{total} = 1 + \frac{1}{100} P_{total}\) Solve for \(P_{total}\): \(P_{total} = \frac{700}{3}\) Substitute the value of \(P_{total}\) back into the first equation: \(x_A = \frac{2}{350} \cdot \frac{700}{3}\) \(x_A = \frac{1}{3}\) So the mole fraction of liquid A in the solution is \(\frac{1}{3}\), or approximately 0.333.

Key concepts

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid form at a specific temperature. It's an essential concept for understanding how liquids evaporate and combine.
One key point is that every liquid has a unique vapor pressure, which increases with temperature.
This is because higher temperatures provide energy to more molecules to escape from the liquid to the vapor phase.

• Vapor pressure helps predict how liquids will behave when combined.
• A higher vapor pressure means the liquid is more volatile and evaporates quickly.

When two volatile liquids are mixed, like in our exercise, their vapor pressures contribute to the total pressure above the solution based on their mole fractions and individual vapor pressures. In this way, Raoult's Law is crucial, as it helps us determine the partial pressures of each component in the solution.

Volatile Liquids

Volatile liquids are substances that evaporate readily at room temperature due to their high vapor pressures. This property is significant when analyzing solutions, especially when applying Raoult's Law.
Let's discuss key features of volatile liquids:

• They tend to have low boiling points.
• Their high vapor pressures make them essential in predicting how solutions will behave under given conditions.

In our example, we dealt with a mixture of two volatile liquids, A and B.
Liquid A had a higher vapor pressure than liquid B, indicating it is more volatile.
This difference in volatility affects how the mole fractions of the liquids will influence the vapor above the solution, and why the solution has a distinct ratio in the vapor phase.

Mole Fraction

Mole fraction is a way to express the concentration of a component in a mixture.
It is defined as the number of moles of a particular component divided by the total moles in the solution.

• Mole fraction is a dimensionless quantity, meaning it has no units.
• The sum of mole fractions for all components in a solution is always equal to 1.

In our exercise, we found that the mole fraction of liquid A in the vapor was double that in the solution. This relationship helped us determine the specific mole fraction of liquid A.
By using Raoult's Law and the conditions given, we calculated the mole fraction of liquid A in the solution to be approximately 0.333, illustrating how these concepts interplay in practical scenarios.