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Two liquids, A and B, are immiscible for \(x_{A}=x_{B}\) \(=0.5,\) for \(T < 75.0^{\circ} \mathrm{C}\) and completely miscible for \(T > 75.0^{\circ} \mathrm{C} .\) Sketch the phase diagram, showing as much information as you can from these observations.

Short Answer

Expert verified
To sketch the phase diagram for immiscible liquids A and B at \(x_A = x_B = 0.5\) and \(T < 75.0^{\circ} \mathrm{C}\), draw temperature (T) on the y-axis and the mole fractions of A and B as \(x_A\) and \(x_B\) on the x-axis. Mark a vertical line at \(x_A = 0.5\) and \(x_B = 0.5\), which separates the immiscible phases for the given temperature range. Mark a horizontal line for \(T = 75.0 ^{\circ} \mathrm{C}\), which separates the miscible and immiscible regions. The region above this line represents complete miscibility for both liquids A and B.

Step by step solution

01

Draw the axes

First, let's draw the axes for the phase diagram. Set temperature (T) on the vertical (y) axis and the mole fractions of A and B as \(x_A\) and \(x_B\) on the horizontal (x) axis.
02

Determine the immiscible region

For \(x_A = x_B = 0.5\) and \(T < 75.0^{\circ} \mathrm{C}\), the liquids are immiscible. Mark a vertical line at \(x_A = 0.5\) and \(x_B = 0.5\) on the phase diagram. This line separates the two immiscible phases for the given temperature range. In this region, we can find two phases; liquid A is on the left side of the vertical line, while liquid B is on the right side.
03

Determine the miscible region

For \(T > 75.0^{\circ} \mathrm{C}\), the liquids are completely miscible. Therefore, mark a horizontal line across the phase diagram, which separates the region above as the miscible region, where phases A and B are completely mixed together.
04

Label the regions

Now, let's label the regions we found in the previous steps. The region below the horizontal line and to the left of the vertical line represents pure liquid A; to the right of the vertical line represents pure liquid B. Above the horizontal line, it represents complete miscibility for both A and B liquids.
05

Complete the diagram

Now, the phase diagram is complete. It shows the immiscible region and the miscible region for the given liquids and they change with temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Immiscible Liquids
In the world of chemistry, not all liquids mix well together. Immiscible liquids are those that do not form a uniform mixture when combined. An easy-to-understand example could be oil and water; when you try to combine them, they stay separated into distinct layers rather than mixing.
The key reason for this behavior lies in the different molecular properties of the liquids involved. Factors such as polarity and molecular size lead to these liquids tending to repel each other instead of mixing. In the original exercise, liquids A and B are described as immiscible under certain conditions, specifically when their mole fractions are at 0.5 and the temperature is below 75.0°C. This means that under these conditions, they are analogous to oil and water inside a glass, each occupying separate spaces.
Mole Fraction
Mole fraction is a concept that helps us understand the composition of a mixture by giving us the ratio of moles of a particular component to the total moles in the mixture. It's expressed as follows:
  • For component A: \(x_A = \frac{n_A}{n_A + n_B}\)
  • For component B: \(x_B = \frac{n_B}{n_A + n_B}\)

In the context of a phase diagram, the mole fraction is plotted on the horizontal axis. The exercise defines a specific mole fraction condition where \(x_A\) and \(x_B\) are both 0.5, meaning in this scenario, the two liquids have equal numbers of moles. Understanding mole fractions is crucial for sketching phase diagrams since it determines how each liquid component interacts under different conditions.
Temperature Dependence
Temperature is a crucial factor affecting the behavior and interactions of liquids. Temperature dependence refers to how the properties of a liquid mixture change as temperature varies. For liquids A and B, the phase diagram indicates a significant dependence on temperature.
Below 75°C, the liquids are immiscible; however, if the temperature increases above 75°C, they become completely miscible. This change illustrates how temperature can alter molecular interactions, often making it possible for liquids to mix fully. As temperature rises, the kinetic energy of molecules increases, overcoming the forces that kept them separate at lower temperatures.
Miscibility
Miscibility is the ability of two substances to form a homogeneous solution when mixed. When two liquids are miscible, they dissolve in each other in any proportion, leading to a consistent and uniform mixture. The exercise at hand indicates that above 75°C, liquids A and B become miscible.
This concept is essential in various fields, from designing chemical processes to formulating pharmaceuticals. In the phase diagram, this is shown where a horizontal line separates the region of complete miscibility from the immiscible region, indicating a critical temperature threshold. This line denotes the transformation from two-phase to single-phase behavior as temperature changes.
Binary Mixture
A binary mixture refers to a system that consists of two different components. In our example with liquids A and B, we're dealing with a binary mixture. Understanding a binary mixture involves looking at how each component behaves individually and how they interact when combined.
In phase diagrams, binary mixtures are represented by the interplay between their components under various conditions such as temperature and concentration. The diagram provides a visual map to anticipate how the mixture behaves under different scenarios. For the given exercise, the binary mixture of A and B isn't fully miscible at all temperatures and compositions, highlighting that not all binary mixtures behave uniformly.

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Most popular questions from this chapter

An ideal solution is made up of the volatile liquids A and \(\mathrm{B}\), for which \(P_{A}^{*}=165\) Torr and \(P_{B}^{*}=85.1\) Torr. As the pressure is reduced, the first vapor is observed at a total pressure of \(110 .\) Torr. Calculate \(x_{\mathrm{A}}\).

At \(-47.0^{\circ} \mathrm{C}\), the vapor pressure of ethyl bromide is 10.0 Torr and that of ethyl chloride is 40.0 Torr. Assume that the solution is ideal. Assume there is only a trace of liquid present and the mole fraction of ethyl chloride in the vapor is 0.80 and answer these questions: a. What is the total pressure and the mole fraction of ethyl chloride in the liquid? b. If there are \(5.00 \mathrm{mol}\) of liquid and \(3.00 \mathrm{mol}\) of vapor present at the same pressure as in part (a), what is the overall composition of the system?

Assume that 1-bromobutane and 1-chlorobutane form an ideal solution. At \(273 \mathrm{K}, P_{\text {chloro }}^{*}=3790 \mathrm{Pa}\) and \(P_{\text {bromo}}^{*}=1394\) Pa. When only a trace of liquid is present at \(273 \mathrm{K}, y_{\text {chloro}}=0.750\). a. Calculate the total pressure above the solution. b. Calculate the mole fraction of 1 -chlorobutane in the solution. c. What value would \(Z_{\text {chloro}}\) have in order for there to be 4.86 mol of liquid and 3.21 mol of gas at a total pressure equal to that in part (a)? [Note: This composition is different from that of part (a).]

An ideal solution is formed by mixing liquids A and \(\mathrm{B}\) at \(298 \mathrm{K} .\) The vapor pressure of pure \(\mathrm{A}\) is 151 Torr and that of pure \(\mathrm{B}\) is 84.3 Torr. If the mole fraction of \(\mathrm{A}\) in the vapor is \(0.610,\) what is the mole fraction of \(A\) in the solution?

The dissolution of 7.75 g of a substance in 825 g of benzene at \(298 \mathrm{K}\) raises the boiling point by \(0.575^{\circ} \mathrm{C}\). Note that \(K_{f}=5.12 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}, K_{b}=2.53 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1},\) and the density of benzene is \(876.6 \mathrm{kg} \mathrm{m}^{-3} .\) Calculate the freezing point depression, the ratio of the vapor pressure above the solution to that of the pure solvent, the osmotic pressure, and the molecular weight of the solute. \(P_{\text {benzene }}^{*}=103\) Torr at \(298 \mathrm{K}\).

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