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Chapter 13: Problem 68

At \(20^{\circ} \mathrm{C}\), the vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(10 \mathrm{kPa}\), and that of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is \(2.9 \mathrm{kPa}\). Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fraction of a solution that has a vapor pressure of \(4.7 \mathrm{kPa}\) at \(20^{\circ} \mathrm{C} ?\) (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?

Short Answer

Expert verified
The composition in mole fraction of the solution is 0.2535 for benzene and 0.7465 for toluene. The mole fraction of benzene in the vapor above the solution is 0.5394.

Step by step solution

01

Raoult's Law for ideal solutions

Raoult's Law states that the partial pressure of a component in an ideal solution is equal to the mole fraction of that component multiplied by its vapor pressure: \(P_{A} = x_{A}P_{A}^{\circ}\). In our problem, component A will be benzene, and component B will be toluene. We are given the vapor pressures of both components, and the total vapor pressure of the solution.
02

Write the expression for the partial pressure of benzene and toluene

Using Raoult's Law, let's write the expressions for the partial pressures of benzene (A) and toluene (B): \(P_{A} = x_{A}P_{A}^{\circ}\) \(P_{B} = x_{B}P_{B}^{\circ}\)
03

Find the mole fractions

We know that the total vapor pressure of the solution is 4.7 kPa: \(P_{total} = P_A + P_B = 4.7\) Since \(x_B = 1 - x_A\), we can substitute the expression for \(P_B\) in terms of \(x_A\): \(4.7 = x_A \cdot 10 + (1 - x_A) \cdot 2.9\) Now, we can solve for the mole fraction of benzene, \(x_A\): \(4.7 = 10x_A + 2.9 - 2.9x_A\) \(1.8 = 7.1x_A\) \(x_A = 0.2535\) Now, we can find the mole fraction of toluene, \(x_B\): \(x_B = 1 - x_A = 1 - 0.2535 = 0.7465\) Thus, the composition in mole fraction of the solution is 0.2535 for benzene and 0.7465 for toluene.
04

Find the mole fraction of benzene in the vapor above the solution

To find the mole fraction of benzene in the vapor, we will use the definition of mole fraction: \(y_A = \frac{P_A}{P_{total}}\) We already know the partial pressure of benzene from the previous steps: \(P_A = x_A P_A^{\circ} = 0.2535 \cdot 10 = 2.535\) Now plug in the values and find the mole fraction of benzene in the vapor: \(y_A = \frac{2.535}{4.7} = 0.5394\) Hence, the mole fraction of benzene in the vapor above the solution is 0.5394.

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Key Concepts

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

Ideal Solutions
Ideal solutions are a special type of liquid mixture where the intermolecular forces between the different types of molecules are similar. This means that the behaviors of individual components in the solution closely mimic how they behave in their pure states. This is why concepts like Raoult's Law hold true for ideal solutions.

In ideal solutions, each component contributes to the overall behavior of the mixture proportionally to its mole fraction. This proportionality allows predictions of properties like vapor pressure, simply by understanding the components:
  • Intermolecular interactions are uniformly dissimilar, allowing simpler mathematical predictions.
  • Components do not change each other's chemical properties or physical state significantly.
  • Raoult's Law is a key tool used to calculate various properties of the mixture.
Understanding these mixtures forms a solid foundation for exploring more complex solutions and non-ideal behavior.
Vapor Pressure
Vapor pressure is a crucial concept in understanding the behavior of solutions in gaseous states. It refers to the pressure exerted by a vapor in equilibrium with its liquid phase. At any given temperature, the vapor pressure is a fixed quantity for a pure substance. This is why vapor pressures are essential data points in exercises involving Raoult's Law.

For instance, Raoult's Law expresses the total vapor pressure of an ideal solution as the sum of the partial pressures exerted by each component. Each of these partial pressures is calculated using the vapor pressure of the pure component and its mole fraction in the solution:
  • Vapor pressure depends on temperature: a key reason why conditions like temperature must be specified in problems.
  • In solutions, each component contributes to the total vapor pressure proportionally to its concentration.
  • This relationship helps predict how mixtures will behave when separated or when interacting with their surrounding environment.
Thus, mastering vapor pressures aids in predicting how substances distribute between phases.
Mole Fraction
Mole fraction is a vital measurement to understand the composition of mixtures. It expresses the ratio of moles of one component to the total moles in the solution. In terms of Raoult's Law, the mole fraction plays a significant role as it directly influences the partial pressures of each component, which sum up to create the total vapor pressure.

The mole fraction of a component, say benzene, in a two-component solution can be found by using:
  • Calculating the total moles of all components in the solution.
  • Dividing the moles of benzene by this total.
  • Understanding these fractions helps predict dynamics like evaporation and boiling within solutions.
Mole fractions are dimensionless, always between 0 and 1, describing the concentration without the need for units. This simplicity allows for smooth calculations and easy insight into solution behavior.

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Most popular questions from this chapter

Compounds like sodium stearate, called "surfactants" in general, can form structures known as micelles in water, once the solution concentration reaches the value known as the critical micelle concentration (cmc). Micelles contain dozens to hundreds of molecules. The cmc depends on the substance, the solvent, and the temperature. At and above the \(\mathrm{cmc}\), the properties of the solution vary drastically. (a) The turbidity (the amount of light scattering) of solutions increases dramatically at the \(\mathrm{cmc}\). Suggest an explanation. (b) The ionic conductivity of the solution dramatically changes at the \(\mathrm{cmc}\). Suggest an explanation. (c) Chemists have developed fluorescent dyes that glow brightly only when the dye molecules are in a hydrophobic environment. Predict how the intensity of such fluorescence would relate to the concentration of sodium stearate as the sodium stearate concentration approaches and then increases past the \(\mathrm{cmc}\).

Some soft drinks contain up to 85 ppm oxygen. (a) What is this concentration in mol/L? (b) What partial pressure of \(\mathrm{O}_{2}\) above water is needed to obtain \(85 \mathrm{ppm} \mathrm{O}_{2}\) in water at \(10^{\circ} \mathrm{C} ?\) (The Henry's law constant for \(\mathrm{O}_{2}\) at this temperature is \(\left.1.69 \times 10^{-5} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa} .\right)\)

At \(35^{\circ} \mathrm{C}\) the vapor pressure of acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO},\) is 47.9 \(\mathrm{kPa}\), and that of carbon disulfide, \(\mathrm{CS}_{2}\), is \(66.7 \mathrm{kPa}\). A solution composed of an equal number of moles of acetone and carbon disulfide has a vapor pressure of \(86.7 \mathrm{kPa}\) at \(35^{\circ} \mathrm{C} .(\mathbf{a})\) What would be the vapor pressure of the solution if it exhibited ideal behavior? (b) Based on the behavior of the solution, predict whether the mixing of acetone and carbon disulfide is an exothermic \(\left(\Delta H_{\text {soln }}<0\right)\) or endothermic \(\left(\Delta H_{\text {soln }}>0\right)\) process.

(a) Do colloids made only of gases exist? Why or why not? (b) In the 1850s, Michael Faraday prepared ruby-red colloids of gold nanoparticles in water that are still stable today. These brightly colored colloids look like solutions. What experiment(s) could you do to determine whether a given colored preparation is a solution or colloid?

(a) Calculate the vapor pressure of water above a solution prepared by adding \(22.5 \mathrm{~g}\) of lactose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) to \(200.0 \mathrm{~g}\) of water at \(338 \mathrm{~K}\). (Vapor-pressure data for water are given in Appendix B.) (b) Calculate the mass of propylene glycol \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{2}\right)\) that must be added to \(0.340 \mathrm{~kg}\) of water to reduce the vapor pressure by \(384 \mathrm{~Pa}\) at \(40^{\circ} \mathrm{C}\).
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