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Chapter 11: Problem 58

A solution is prepared by mixing \(0.0300 \mathrm{~mol} \mathrm{CH}_{2} \mathrm{Cl}_{2}\) and \(0.0500\) \(\mathrm{mol} \mathrm{CH}_{2} \mathrm{Br}_{2}\) at \(25^{\circ} \mathrm{C}\). Assuming the solution is ideal, calculate the composition of the vapor (in terms of mole fractions) at \(25^{\circ} \mathrm{C}\). At \(25^{\circ} \mathrm{C}\), the vapor pressures of pure \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and pure \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) are 133 and \(11.4\) torr, respectively.

Short Answer

Expert verified
The composition of the vapor at 25掳C is: Mole fraction of CH鈧侰l鈧 (Y鈧) = 0.875 Mole fraction of CH鈧侭r鈧 (Y鈧) = 0.125

Step by step solution

01

Calculate the mole fractions of CH鈧侰l鈧 and CH鈧侭r鈧 in the solution.

First, we need to find the mole fractions of CH鈧侰l鈧 and CH鈧侭r鈧 in the solution. To do this, we divide the moles of each component by the total moles in the solution. Total moles = moles of CH鈧侰l鈧 + moles of CH鈧侭r鈧 Total moles = 0.0300 mol + 0.0500 mol = 0.0800 mol Now, we find the mole fractions: Mole fraction of CH鈧侰l鈧 (X鈧) = (moles of CH鈧侰l鈧) / (total moles) = 0.0300 mol / 0.0800 mol = 0.375 Mole fraction of CH鈧侭r鈧 (X鈧) = (moles of CH鈧侭r鈧) / (total moles) = 0.0500 mol / 0.0800 mol = 0.625
02

Calculate the partial pressures of CH鈧侰l鈧 and CH鈧侭r鈧.

Next, we will use Raoult's Law to calculate the partial pressures (P鈧 and P鈧) of CH鈧侰l鈧 and CH鈧侭r鈧 in the vapor phase. Raoult's Law states: P_i = X_i * P_i掳 where P_i is the partial pressure of component i, X_i is the mole fraction of component i in the solution, and P_i掳 is the vapor pressure of the pure component i. For CH鈧侰l鈧 (component 1): P鈧 = X鈧 * P鈧伮 = 0.375 * 133 torr = 49.875 torr For CH鈧侭r鈧 (component 2): P鈧 = X鈧 * P鈧偮 = 0.625 * 11.4 torr = 7.125 torr
03

Calculate the total pressure of the vapor phase.

Now we will find the total pressure (P_total) of the vapor phase by adding the partial pressures of each component: P_total = P鈧 + P鈧 = 49.875 torr + 7.125 torr = 57.000 torr
04

Calculate the mole fractions of CH鈧侰l鈧 and CH鈧侭r鈧 in the vapor phase.

Finally, we will use the partial pressures to calculate the mole fractions of each component in the vapor phase: Mole fraction of CH鈧侰l鈧 in vapor (Y鈧) = P鈧 / P_total = 49.875 torr / 57.000 torr = 0.875 Mole fraction of CH鈧侭r鈧 in vapor (Y鈧) = P鈧 / P_total = 7.125 torr / 57.000 torr = 0.125 The composition of the vapor at 25掳C is: Mole fraction of CH鈧侰l鈧 (Y鈧) = 0.875 Mole fraction of CH鈧侭r鈧 (Y鈧) = 0.125

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Key Concepts

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

Ideal Solution
An ideal solution is a theoretical concept where the mixing of components does not result in any volume change or heat exchange. This means that the interactions between unlike molecules are similar to those between like molecules. In other words, the intermolecular forces in the mixture are essentially uniform.

Raoult's Law is often applicable to ideal solutions. When two liquids form an ideal solution, each liquid behaves as if it were in its pure form but reduced by its mole fraction.
  • No expansion or contraction occurs upon mixing.
  • The solution displays no heat of mixing. This means there is no absorption or release of energy.
  • Raoult's Law holds true strictly.
In our exercise, the solution of dichloromethane (CH鈧侰l鈧) and dibromomethane (CH鈧侭r鈧) is assumed to be ideal, hence Raoult's Law can be effectively applied.
Partial Pressure
To understand partial pressure, imagine a gas mixture where each gas exerts a pressure independently. The partial pressure of any component is the pressure it would exert if it occupied the entire volume alone.

Raoult's Law helps us calculate this in an ideal solution: \[ P_i = X_i \cdot P_i^\circ \]where:
  • \( P_i \) is the partial pressure of component \( i \).
  • \( X_i \) is the mole fraction of component \( i \) in the solution.
  • \( P_i^\circ \) is the vapor pressure of the pure component \( i \).
In the solution involving CH鈧侰l鈧 and CH鈧侭r鈧, partial pressures were calculated for each component. These values depend on each component's mole fraction in the solution and their respective vapor pressures in their pure states.
Vapor Pressure
Vapor pressure is the pressure at which a liquid's molecules are in equilibrium in the gaseous state. For any liquid, vapor pressure signifies how volatile the liquid is.

A higher vapor pressure means a higher tendency for the molecules to escape into the gas phase.
  • Pure CH鈧侰l鈧 with a vapor pressure of 133 torr indicates more molecules escapable to the vapor phase compared to CH鈧侭r鈧 with only 11.4 torr.
  • The difference in vapor pressures reflects different volatilities between the two substances.
In our solution scenario, these values are used to compute the component's partial pressures via Raoult's Law, integrating directly into the understanding of how the vapor phase will form over the solution mixture.
Mole Fraction
Mole fraction is a way of expressing the concentration of a component in a mixture. It is calculated as the ratio of the moles of a component to the total moles present. This dimensionless quantity plays a critical role in applying Raoult's Law.

Formulaically, for any substance \( A \) in a mixture:\[ X_A = \frac{\text{moles of } A}{\text{total moles in solution}} \]
  • The mole fraction of CH鈧侰l鈧 in the solution was found to be 0.375, indicating that 37.5% of the total moles in the solution are CH鈧侰l鈧.
  • Conversely, the mole fraction of CH鈧侭r鈧 was 0.625, signifying that 62.5% are CH鈧侭r鈧.
Mole fractions are pivotal for determining the partial pressures of each component in an ideal solution, which further allows us to determine their respective compositions in the vapor phase as seen in the exercise.

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Most popular questions from this chapter

A solution of phosphoric acid was made by dissolving \(10.0 \mathrm{~g}\) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in \(100.0 \mathrm{~mL}\) water. The resulting volume was \(104 \mathrm{~mL}\) Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\).

An aqueous antifreeze solution is \(40.0 \%\) ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) by mass. The density of the solution is \(1.05 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the molality, molarity, and mole fraction of the ethylene glycol.

In flushing and cleaning columns used in liquid chromatography to remove adsorbed contaminants, a series of solvents is used. Hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\), chloroform \(\left(\mathrm{CHCl}_{3}\right)\), methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\), and water are passed through the column in that order. Rationalize the order in terms of intermolecular forces and the mutual solu- bility (miscibility) of the solvents.

Erythrocytes are red blood cells containing hemoglobin. In a saline solution they shrivel when the salt concentration is high and swell when the salt concentration is low. In a \(25^{\circ} \mathrm{C}\) aqueous solution of \(\mathrm{NaCl}\), whose freezing point is \(-0.406^{\circ} \mathrm{C}\), erythrocytes neither swell nor shrink. If we want to calculate the osmotic pressure of the solution inside the erythrocytes under these conditions, what do we need to assume? Why? Estimate how good (or poor) of an assumption this is. Make this assumption and calculate the osmotic pressure of the solution inside the erythrocytes.

Explain the terms isotonic solution, crenation, and hemolysis.
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