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The U.S. Food and Drug Administration lists dichloromethane \(\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)\) and carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) among the many cancer-causing chlorinated organic compounds. What are the partial pressures of these substances in the vapor above a solution of \(1.60 \mathrm{~mol}\) of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and \(1.10 \mathrm{~mol}\) of \(\mathrm{CCl}_{4}\) at \(23.5^{\circ} \mathrm{C}\) ? The vapor pressures of pure \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and \(\mathrm{CCl}_{4}\) at \(23.5^{\circ} \mathrm{C}\) are 352 torr and 118 torr, respectively. (Assume ideal behavior.)

Short Answer

Expert verified
Partial pressures: \(\text{CH}_2\text{Cl}_2\) ≈ 208.15 torr, \(\text{CCl}_4\) ≈ 48.04 torr.

Step by step solution

01

Identify Given Information

The problem provides the following data:- Moles of \(\text{CH}_2\text{Cl}_2\): 1.60 mol- Moles of \(\text{CCl}_4\): 1.10 mol- Vapor pressure of pure \(\text{CH}_2\text{Cl}_2\) (\text{P\textsubscript{CH\textsubscript{2}Cl\textsubscript{2}}}\textsubscript{0}}): 352 torr- Vapor pressure of pure \(\text{CCl}_4\) (\text{P\textsubscript{CCl\textsubscript{4}}}\textsubscript{0}}): 118 torr
02

Calculate Mole Fractions

The mole fraction of a component in a solution is given by \(\frac{\text{moles of component}}{\text{total moles}} \). Using this formula for each component, calculate the mole fractions:- Total moles = 1.60 + 1.10 = 2.70 mol- Mole fraction of \(\text{CH}_2\text{Cl}_2\) (\( \text{x}_{\text{CH}_2\text{Cl}_2} \)) = \(\frac{1.60}{2.70} \)- Mole fraction of \(\text{CCl}_4\) (\( \text{x}_{\text{CCl}_4} \)) = \(\frac{1.10}{2.70} \)
03

Calculate Individual Partial Pressures

The partial pressure of each component in the vapor above the solution is given by Raoult's Law: \(\text{P}_{\text{component}} = \text{x}_{\text{component}} \times \text{P}_{\text{component0}} \).- For \(\text{CH}_2\text{Cl}_2\): \(\text{P}_{\text{CH}_2\text{Cl}_2} = \frac{1.60}{2.70} \times 352 \)- For \(\text{CCl}_4\): \(\text{P}_{\text{CCl}_4} = \frac{1.10}{2.70} \times 118 \)
04

Perform Calculations

Perform the calculations for each partial pressure:- \(\text{P}_{\text{CH}_2\text{Cl}_2} = \frac{1.60}{2.70} \times 352 \approx 208.15 \text{ torr} \)- \(\text{P}_{\text{CCl}_4} = \frac{1.10}{2.70} \times 118 \approx 48.04 \text{ torr} \)

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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 essential to understanding how the vapor pressure in a solution behaves. The law states that the partial pressure of each component in a liquid solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. This can be expressed mathematically as: \( P_{\text{component}} = x_{\text{component}} \times P_{\text{component0}} \) where \( P_{\text{component}} \) is the partial pressure of the component, \( x_{\text{component}} \) is the mole fraction, and \( P_{\text{component0}} \) is the vapor pressure of the pure component.

This law is particularly useful when dealing with ideal solutions, where the interactions between different molecules are similar to the interactions between molecules of the same substance.

In our exercise, Raoult's Law helps to find the partial pressures of dichloromethane (CHâ‚‚Clâ‚‚) and carbon tetrachloride (CClâ‚„) by taking into account their mole fractions and pure component vapor pressures at a given temperature.
Vapor pressure
Vapor pressure is a measure of the tendency of a liquid's molecules to escape into the vapor phase. Each pure substance has a unique vapor pressure at a given temperature, which reflects how easily its molecules can transition from liquid to vapor.

Understanding vapor pressure is crucial when working with solutions because it directly affects the mixture's behavior. In an ideal solution, the total vapor pressure is the sum of the partial pressures of the individual components.

In our example, the vapor pressure of pure dichloromethane (CH₂Cl₂) at 23.5°C is 352 torr, and the vapor pressure of pure carbon tetrachloride (CCl₄) at the same temperature is 118 torr. These values are used to determine the partial pressures of each component in the vapor phase above the solution.
Mole fraction
The mole fraction is a way of expressing the concentration of a component in a solution. It is defined as the ratio of the number of moles of a given component to the total number of moles of all components in the solution. Mathematically, it is given by: \( x_{\text{component}} = \frac{\text{moles of component}}{\text{total moles}} \).

Mole fractions are crucial when applying Raoult's Law to calculate partial pressures. In our problem, we calculate the mole fractions of dichloromethane (CHâ‚‚Clâ‚‚) and carbon tetrachloride (CClâ‚„) as follows:
- Total moles in the solution: 1.60 mol CHâ‚‚Clâ‚‚ + 1.10 mol CClâ‚„ = 2.70 mol
- Mole fraction of CHâ‚‚Clâ‚‚ (\( x_{\text{CHâ‚‚Clâ‚‚}} \)) = \( \frac{1.60}{2.70} \)
- Mole fraction of CClâ‚„ (\( x_{\text{CClâ‚„}} \)) = \( \frac{1.10}{2.70} \)

These mole fractions are then used to calculate the partial pressures of each component in the vapor phase.
Ideal behavior
Ideal behavior in the context of solutions means that the interactions between different molecules are similar to the interactions between molecules of the same substance. This assumption allows us to use Raoult's Law directly without corrections for deviations.

An ideal solution obeys Raoult's Law perfectly, and the partial pressures can be calculated straightforwardly based on the mole fractions and the vapor pressures of the pure components.

In our exercise, the assumption of ideal behavior simplifies the calculations. This assumption is valid because we are dealing with a simple binary mixture of CHâ‚‚Clâ‚‚ and CClâ‚„, both of which are similar types of organic compounds. Their interactions are not significantly different from those in their pure states, making the ideal behavior assumption reasonable.

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Most popular questions from this chapter

A solution of \(1.50 \mathrm{~g}\) of solute dissolved in \(25.0 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C}\) has a boiling point of \(100.45^{\circ} \mathrm{C}\). (a) What is the molar mass of the solute if it is a nonvolatile nonelectrolyte and the solution behaves ideally \(\left(d\right.\) of \(\mathrm{H}_{2} \mathrm{O}\) at \(\left.25^{\circ} \mathrm{C}=0.997 \mathrm{~g} / \mathrm{mL}\right) ?\) (b) Conductivity measurements show that the solute is ionic with general formula \(\mathrm{AB}_{2}\) or \(\mathrm{A}_{2} \mathrm{~B}\). What is the molar mass if the solution behaves ideally? (c) Analysis indicates that the solute has an empirical formula of \(\mathrm{CaN}_{2} \mathrm{O}_{6}\). Explain the difference between the actual formula mass and that calculated from the boiling point elevation. (d) Find the van't Hoff factor ( \(i\) ) for this solution.

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