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At 303. K, the vapor pressure of benzene is 120. Torr and that of hexane is 189 Torr. Calculate the vapor pressure of a solution for which \(x_{\text {benzene }}=0.28\) assuming ideal behavior.

Short Answer

Expert verified
The vapor pressure of the solution with a mole fraction of benzene of 0.28 at 303 K, assuming ideal behavior, is 169.68 Torr.

Step by step solution

01

Identify known quantities

The given quantities in the exercise are: 1. Vapor pressure of benzene (\(P^0_{benzene}\)) = 120 Torr 2. Vapor pressure of hexane (\(P^0_{hexane}\)) = 189 Torr 3. Mole fraction of benzene (\(x_{benzene}\)) = 0.28
02

Calculate mole fraction of hexane

To calculate the mole fraction of hexane (\(x_{hexane}\)), we can use the fact that the sum of the mole fractions of all components in the solution is always equal to 1: \[x_{hexane} = 1 - x_{benzene}\] \[x_{hexane} = 1 - 0.28 = 0.72\]
03

Apply Raoult's Law for Benzene

We can now apply Raoult's Law to find the partial pressure of benzene (\(P_{benzene}\)) in the solution. Raoult's Law states that the partial pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapor pressure in the pure state: \[P_{benzene} = x_{benzene} \times P^0_{benzene}\] \[P_{benzene} = 0.28 \times 120 = 33.6 \text{ Torr}\]
04

Apply Raoult's Law for Hexane

Next, we can apply Raoult's Law to find the partial pressure of hexane (\(P_{hexane}\)) in the solution: \[P_{hexane} = x_{hexane} \times P^0_{hexane}\] \[P_{hexane} = 0.72 \times 189 = 136.08 \text{ Torr}\]
05

Find the vapor pressure of the solution

The vapor pressure of the solution is the sum of the partial pressures of its components: \[P_{solution} = P_{benzene} + P_{hexane}\] \[P_{solution} = 33.6 \text{ Torr} + 136.08 \text{ Torr} = 169.68 \text{ Torr}\] The vapor pressure of the solution is 169.68 Torr.

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Key Concepts

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

Vapor Pressure
Understanding vapor pressure is essential when studying the properties of liquids and solutions. It refers to the pressure exerted by a vapor in equilibrium with its liquid or solid form at a given temperature. When a liquid is placed in a closed container, some molecules at the surface evaporate into the vapor phase. Over time, an equilibrium is reached where the rate of evaporation equals the rate of condensation of the vapor back into the liquid.

This balance creates a steady pressure within the container, known as the vapor pressure, which is dependent on the substance and temperature. The higher the temperature, the more molecules have sufficient kinetic energy to escape into the vapor phase, increasing the vapor pressure. For instance, at 303 K, pure benzene has a vapor pressure of 120 Torr, while hexane has a higher vapor pressure of 189 Torr, indicating that hexane tends to vaporize more readily than benzene at the same temperature.
Mole Fraction
The mole fraction is a way of expressing the concentration of a component in a mixture or a solution. It is defined as the ratio of the number of moles of a particular substance to the total number of moles of all substances present. Represented by the symbol 'x', mole fraction is a unitless quantity that ranges from 0 to 1.

To calculate the mole fraction of a component, you divide the number of moles of that component by the total moles in the mixture. For example, if a solution has a mole fraction of benzene \(x_{\text{benzene}}\) equal to 0.28, it means that benzene makes up 28% of the total moles in the solution. Since the sum of mole fractions in a solution is always 1, the mole fraction of hexane \(x_{\text{hexane}}\) would be 1 minus the mole fraction of benzene, thus 0.72 or 72% of the total moles.
Partial Pressure
Partial pressure is the pressure that a single component in a mixture of gases would exert if it occupied the entire volume by itself at the same temperature. Each gas in a mixture exerts pressure independently of the others, and the total pressure is the sum of all the individual partial pressures. According to Dalton's Law of Partial Pressures, this holds true for ideal gas mixtures.

In the context of solutions and Raoult's Law, the partial pressure of a liquid component is proportional to its vapor pressure when pure and its mole fraction in the solution. For instance, the calculation of the partial pressure of benzene in a solution using Raoult's Law involves multiplying the mole fraction of benzene by its pure vapor pressure. Similarly, the partial pressure of hexane is determined by multiplying its mole fraction by its pure vapor pressure.
Ideal Solutions
An ideal solution is a mixture that obeys Raoult's Law at all concentrations for both solvents and all conditions of temperature and pressure. It assumes that the interactions between molecules of different components are the same as the interactions between molecules of the same component. In practice, this means that the properties of the solution, such as vapor pressure, can be predicted from the properties of the pure components using simple linear relationships.

For example, Raoult's Law states that the vapor pressure of an ideal solution is dependent on the vapor pressures of each component in their pure states and their respective mole fractions in the solution. The assumption of ideal behavior allows us to calculate the partial pressures and thus the total vapor pressure of a solution, as demonstrated in the exercise involving benzene and hexane.

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Most popular questions from this chapter

\(A\) and \(B\) form an ideal solution at \(298 \mathrm{K},\) with \(x_{A}=0.320, P_{A}^{*}=84.3\) Torr, and \(P_{B}^{*}=41.2\) Torr. a. Calculate the partial pressures of \(A\) and \(B\) in the gas phase. b. A portion of the gas phase is removed and condensed in a separate container. Calculate the partial pressures of A and \(\mathrm{B}\) in equilibrium with this liquid sample at \(298 \mathrm{K}\).

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?

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.

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).]

The densities of pure water and ethanol are 997 and \(789 \mathrm{kg} \mathrm{m}^{-3},\) respectively. For \(x_{\text {ethanol }}=0.35,\) the partial molar volumes of ethanol and water are 55.2 and \(17.8 \times 10^{-3} \mathrm{L} \mathrm{mol}^{-1},\) respectively. Calculate the change in volume relative to the pure components when \(2.50 \mathrm{L}\) of a solution with \(x_{\text {ethanol }}=0.35\) is prepared.

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