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A liquid solution consists of \(0.25\) mole fraction ethylene dibromide, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2}\), and \(0.75 \mathrm{~mole}\) fraction propylene dibromide, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Br}_{2} .\) Both ethylene dibromide and propylene dibromide are volatile liquids; their vapor pressures at \(85^{\circ} \mathrm{C}\) are \(173 \mathrm{mmHg}\) and \(127 \mathrm{mmHg}\), respectively. Assume that each compound follows Raoult's law in the solution. Calculate the total vapor pressure of the solution.

Short Answer

Expert verified
The total vapor pressure of the solution is 138.5 mmHg.

Step by step solution

01

Understanding Raoult's Law

Raoult's Law states that the partial vapor pressure of an individual component in a solution is equal to the product of the mole fraction of the component in the liquid phase and its vapor pressure as a pure substance. Mathematically, this can be represented as: \[ P_i = x_i P^0_i \] where \( P_i \) is the partial vapor pressure of the component, \( x_i \) is the mole fraction, and \( P^0_i \) is the vapor pressure of the pure component.
02

Calculate Partial Vapor Pressure for Ethylene Dibromide

Using Raoult's Law, calculate the partial vapor pressure for ethylene dibromide:\[ x_{ ext{ethylene}} = 0.25 \]\[ P^0_{ ext{ethylene}} = 173 \text{ mmHg} \]Now, calculate:\[ P_{ ext{ethylene}} = x_{ ext{ethylene}} imes P^0_{ ext{ethylene}} \]\[ P_{ ext{ethylene}} = 0.25 imes 173 = 43.25 \text{ mmHg} \]
03

Calculate Partial Vapor Pressure for Propylene Dibromide

Similarly, calculate the partial vapor pressure for propylene dibromide:\[ x_{ ext{propylene}} = 0.75 \]\[ P^0_{ ext{propylene}} = 127 \text{ mmHg} \]Now, calculate:\[ P_{ ext{propylene}} = x_{ ext{propylene}} imes P^0_{ ext{propylene}} \]\[ P_{ ext{propylene}} = 0.75 imes 127 = 95.25 \text{ mmHg} \]
04

Calculate the Total Vapor Pressure of the Solution

According to Raoult's Law, the total vapor pressure of the solution is the sum of the partial vapor pressures of the individual components:\[ P_{ ext{total}} = P_{ ext{ethylene}} + P_{ ext{propylene}} \]\[ P_{ ext{total}} = 43.25 + 95.25 = 138.5 \text{ mmHg} \]

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Key Concepts

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

Vapor Pressure Calculation
Understanding how to calculate vapor pressure is key when dealing with mixtures of volatile liquids. Raoult's Law provides the foundation for these calculations. It suggests that the contribution of each liquid to the total vapor pressure of a solution can be determined through its mole fraction and its vapor pressure as a pure substance. By knowing these two components, it's possible to figure out how much vapor pressure each liquid contributes to the entire solution. This is essential when you need to determine how a liquid mix behaves, especially when dealing with boiling points or solvent separation in chemical processes. These calculations rely heavily on using accurate mole fractions and knowing each component's vapor pressure when pure to ensure the results are reflective of real-world behavior. Moreover, the condition that both substances in the mixture behave according to Raoult's Law is critical, as deviations can lead to disparities in the expected vapor pressure.
Mole Fraction
The mole fraction is a crucial concept in the chemistry of solutions. It measures the proportion of that component compared to the total amount of all components in the mixture. Calculated by dividing the number of moles of a particular component by the total number of moles of all substances combined, the mole fraction offers a simple yet effective method of expressing the concentration of each component. For a solution composed of volatile liquids such as ethylene dibromide and propylene dibromide, the mole fraction determines how much influence each liquid has over the total vapor pressure. For instance, in our example, ethylene dibromide has a mole fraction of 0.25, meaning it represents 25% of the total mixture, while propylene dibromide has a mole fraction of 0.75, representing 75% of the mixture. This concept is straightforward but vital when applying Raoult's Law, as precise mole fractions ensure accurate vapor pressure calculations.
Volatile Liquids
Volatile liquids are substances with a high tendency to evaporate or vaporize under standard conditions. This property makes them significant in calculating vapor pressure and assessing how solutions behave. These liquids have significant implications in both industrial applications and everyday chemistry, affecting processes such as distillation and the behavior of mixtures. Volatile liquids have measurable vapor pressures, indicating how rapidly they transition from a liquid state to vapor. In a mixture, each volatile component contributes to the overall vapor pressure proportionately based on its mole fraction. Their volatility dictates their behavior in a solution governed by Raoult's Law, directly affecting the calculation of total vapor pressures. Understanding the nature of volatile liquids allows chemists to predict the behavior of these solutions concerning changes in temperature and pressure.
Partial Vapor Pressure
Partial vapor pressure is the pressure contribution of a single component within a mixture of gases or liquids. In the context of Raoult's Law, it's derived from the product of the component's mole fraction in the liquid phase and the vapor pressure of the pure component. This concept matters because each component in a mixture exerts its vapor pressure independent of others, and the sum of these partial pressures gives the total vapor pressure. For example, in our ethylene and propylene dibromide solution, each chemical's partial pressure could be calculated separately by multiplying its mole fraction by its pure vapor pressure. Partial vapor pressures are a practical measure in many fields, indicating how individual components influence the total pressure and providing insights into the solution's thermodynamic properties. Knowing these helps in adjusting methods such as separation processes or designing solutions to behave in a needed way under different pressures and temperatures.

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Most popular questions from this chapter

Vanillin, \(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3}\), occurs naturally in vanilla extract and is used as a flavoring agent. A 39.1-mg sample of vanillin was dissolved in \(168.5 \mathrm{mg}\) of diphenyl ether, \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \mathrm{O} .\) What is the molality of vanillin in the solution?

Consider two hypothetical pure substances, \(\mathrm{AB}(s)\) and \(\mathrm{XY}(s) .\) When equal molar amounts of these substances are placed in separate \(500-\mathrm{mL}\) samples of water, they undergo the following reactions: $$\begin{gathered}\mathrm{AB}(s) \longrightarrow \mathrm{A}^{+}(a q)+\mathrm{B}^{-}(a q) \\ \mathrm{XY}(s) \longrightarrow \mathrm{XY}(a q) \end{gathered}$$ a. Which solution would you expect to have the lower boiling point? Why? b. Would you expect the vapor pressures of the two solutions to be equal? If not, which one would you expect to have the higher vapor pressure? C. Describe a procedure that would make the two solutions have the same boiling point. d. If you took \(250 \mathrm{~mL}\) of the \(\mathrm{AB}(a q)\) solution prepared above, would it have the same boiling point as the original solution? Be sure to explain your answer. e. The container of \(X Y(a q)\) is left out on the bench top for several days, which allows some of the water to evaporate from the solution. How would the melting point of this solution compare to the melting point of the original solution?

Analysis of a compound gave \(39.50 \%\) C, \(2.21 \% \mathrm{H}\), and \(58.30 \%\) Cl. When \(0.855 \mathrm{~g}\) of this solid was dissolved in \(7.50 \mathrm{~g}\) of naphthalene, the solution had a freezing point of \(78.0^{\circ} \mathrm{C}\). The pure solvent freezes at \(80.0^{\circ} \mathrm{C} ;\) its molal freezing point constant is \(6.8^{\circ} \mathrm{C} / \mathrm{m}\) a. What is the molecular formula of the compound? b. What is its molecular mass to the nearest \(0.1 \mathrm{~g}\) ?

Explain why a carbonated beverage must be stored in a closed container.

Oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), occurs as the potassium or calcium salt in many plants, including rhubarb and spinach. An aqueous solution of oxalic acid is \(0.580 \mathrm{~m} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} .\) The density of the solution is \(1.022 \mathrm{~g} / \mathrm{mL}\). What is the molar concentration?

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