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Chapter 12: Problem 94

The vapor pressure of trichloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3}\right)\) is 100 torr at \(20.0^{\circ} \mathrm{C}\). What is the vapor pressure of trichloroethane, at \(20.0^{\circ} \mathrm{C},\) above a solution containing \(2.00 \mathrm{~g}\) ferrocene \(\left[\mathrm{Fe}\left(\mathrm{C}_{5} \mathrm{H}_{5}\right)_{2},\right.\) which is nonvolatile \(]\) in \(25.0 \mathrm{~g}\) trichloroethane?

Short Answer

Expert verified
The vapor pressure of trichloroethane above the solution is 93.8 torr.

Step by step solution

01

Calculate the moles of solute and solvent

First, we need to calculate the number of moles of the solute (ferrocene) and the solvent (trichloroethane). The molar mass of ferrocene \([\text{Fe}(\text{C}_5\text{H}_5)_2]\) is calculated as follows:- Fe: 1 atom \(\times 55.85\, \text{g/mol} = 55.85\, \text{g/mol}\)- C: 10 atoms \(\times 12.01\, \text{g/mol} = 120.1\, \text{g/mol}\)- H: 10 atoms \(\times 1.01\, \text{g/mol} = 10.1\, \text{g/mol}\)Adding these gives a total molar mass of ferrocene: \(\approx 186.0 \, \text{g/mol}\).The moles of ferrocene (solute) is:\[ n_\text{ferrocene} = \frac{2.00 \, \text{g}}{186.0 \, \text{g/mol}} \approx 0.0108 \, \text{mol} \]The molar mass of trichloroethane \((\text{C}_2\text{H}_3\text{Cl}_3)\) is:- C: 2 atoms \(\times 12.01\, \text{g/mol} = 24.02\, \text{g/mol}\)- H: 3 atoms \(\times 1.01\, \text{g/mol} = 3.03\, \text{g/mol}\)- Cl: 3 atoms \(\times 35.45\, \text{g/mol} = 106.35\, \text{g/mol}\)Adding these gives a total molar mass of trichloroethane: \(\approx 153.4 \, \text{g/mol}\).The moles of trichloroethane (solvent) is:\[ n_\text{trichloroethane} = \frac{25.0 \, \text{g}}{153.4 \, \text{g/mol}} \approx 0.163 \text{mol} \]
02

Calculate the mole fraction of the solvent

The mole fraction of the solvent \((\chi_{\text{trichloroethane}})\) in the solution is given by the formula:\[\chi_{\text{trichloroethane}} = \frac{\text{moles of trichloroethane}}{\text{total moles}} = \frac{0.163}{0.163 + 0.0108} \approx 0.938\]This mole fraction value indicates the proportion of trichloroethane molecules in the solution.
03

Calculate the vapor pressure of the solution using Raoult's Law

Raoult's Law states that the vapor pressure of the solution \(P_{\text{solution}}\) is the product of the mole fraction of the solvent and the vapor pressure of the pure solvent. Thus:\[P_{\text{solution}} = \chi_{\text{trichloroethane}} \times P^{\circ}_{\text{trichloroethane}}\]where \(P^{\circ}_{\text{trichloroethane}} = 100 \, \text{torr}\).So,\[P_{\text{solution}} = 0.938 \times 100 \, \text{torr} = 93.8 \, \text{torr}\]
04

Conclusion

By applying these calculations, we find that the vapor pressure of trichloroethane above the solution is approximately \(93.8 \, \text{torr}\). This decrease in vapor pressure is due to the introduction of the nonvolatile solute, ferrocene, into the solvent.

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Key Concepts

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

Raoult's Law
Raoult's Law is an essential principle in solution chemistry that relates to vapor pressure. It explains how the presence of a solute affects the vapor pressure of a solvent. This law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution. In mathematical terms, it can be expressed as:
\[P_{ ext{solution}} = \chi_{ ext{solvent}} \times P^{\circ}_{ ext{solvent}}\]where:
  • \( P_{\text{solution}} \) is the vapor pressure above the solution.
  • \( \chi_{\text{solvent}} \) is the mole fraction of the solvent.
  • \( P^{\circ}_{\text{solvent}} \) is the vapor pressure of the pure solvent.
Raoult's Law helps us understand how solute particles interact with solvent molecules to lower the solvent's ability to escape into the vapor phase, effectively decreasing its vapor pressure. This is particularly useful when calculating the impact of nonvolatile solutes, as they don't contribute additional vapor pressure themselves.
Mole Fraction
The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the number of moles of a particular component to the total number of moles of all components in the mixture. In the context of our exercise, we specifically look at the mole fraction of the solvent (trichloroethane) in the solution. The formula to calculate the mole fraction \( \chi \) of a component is:
\[\chi = \frac{n_{\text{component}}}{n_{\text{total}}}\]where:
  • \( n_{\text{component}} \) is the number of moles of the component of interest (either solute or solvent).
  • \( n_{\text{total}} \) is the total number of moles in the solution.
Understanding the mole fraction is important because it informs us about the proportion of each substance in the mixture, which directly influences properties like vapor pressure under Raoult's Law.
Solution Chemistry
Solution chemistry revolves around the study of chemical processes that occur in solutions. One key aspect of this field is understanding how solutes and solvents interact. Solutions form when a solute is dissolved in a solvent, resulting in a homogeneous mixture. The nature of each component鈥攚hether volatile or nonvolatile鈥攁ffects physical properties like boiling point, freezing point, and vapor pressure.
In our exercise, trichloroethane acts as the solvent, and ferrocene is the solute. Since ferrocene is nonvolatile, it doesn't add to the vapor phase. Instead, it occupies space between solvent molecules, reducing the number of solvent molecules that can escape as vapor. This phenomenon is a practical insight into colligative properties, which depend solely on the number of solute particles rather than their identity.
Understanding these concepts helps us predict and explain the behavior of solutions in various chemical processes and real-world applications, such as the formulation of products like pharmaceuticals, perfumes, and antifreeze.

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Most popular questions from this chapter

We analyzed the enthalpy change that accompanies the formation of a solution from the pure solute and solvent by considering three changes that must occur: (1) separate the solvent molecules; (2) separate the solute molecules; and (3) bring the solute particles and solvent particles together. If the enthalpy of solution is negative (an exothermic process), what can be said about the relative enthalpy changes of the three processes just listed?

The solubility of ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) in water at \(20^{\circ} \mathrm{C}\) and 0.300 atm pressure is \(1.27 \times 10^{-4}\) molal. (a) Calculate the Henry's law constant for this gas in units of molal/torr. (b) How many grams of ethylene are dissolved in \(1.00 \mathrm{~kg}\) water at \(20^{\circ} \mathrm{C}\) if the pressure of the gas is 500 torr?

List and define the colligative properties, and give the units used for concentrations for each.

Give quantitative directions for preparing \(15.5 \mathrm{~g}\) of a \(1.00 \%\) solution of boric acid \(\left(\mathrm{H}_{3} \mathrm{BO}_{3}\right)\). What is the molal concentration of this solution?

At a pressure of 760 torr, acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH} ;\right.\) boiling point \(=118.1^{\circ} \mathrm{C}\) ) and 1,1 -dibromoethane \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2} ;\right.\) boiling point \(\left.=109.5^{\circ} \mathrm{C}\right)\) form an azeotropic mixture, boiling at \(103.7^{\circ} \mathrm{C}\), that is \(25 \%\) by mass acetic acid. At the boiling point of the azeotrope \(\left(103.7^{\circ} \mathrm{C}\right),\) the vapor pressure of pure acetic acid is 471 torr and that of pure 1,1 -dibromoethane is 637 torr. (a) Calculate the vapor pressure of each component and the total vapor pressure at \(103.7^{\circ} \mathrm{C}\) if the solution had obeyed Raoult's law for both components. (b) Compare your answer to part a with the actual vapor pressure of the azeotrope. Is the deviation of this azeotropic mixture from Raoult's law positive or negative? (c) Compare the attractive forces between acetic acid and dibromoethane with those in the two pure substances.
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