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Dry nitrogen gas is bubbled through liquid benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) at \(20.0^{\circ} \mathrm{C} .\) From \(100.0 \mathrm{L}\) of the gaseous mixture of nitrogen and benzene, \(24.7 \mathrm{g}\) benzene is condensed by passing the mixture through a trap at a temperature where nitrogen is gaseous and the vapor pressure of benzene is negligible. What is the vapor pressure of benzene at \(20.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The vapor pressure of benzene at 20.0°C is 0.9785 atm.

Step by step solution

01

Understand the given information

We are given the following information: 1. Initial volume of the gaseous mixture of nitrogen and benzene: 100.0 L 2. Mass of benzene condensed: 24.7 g 3. Temperature of the system: 20.0°C = 293.15 K (converting to Kelvin) We need to determine the vapor pressure of benzene at this temperature.
02

Calculate moles of condensed benzene

First, we need to calculate the moles of condensed benzene using the given mass and the molar mass of benzene (C6H6): Molar mass of benzene = (6 × 12.01 g/mol C) + (6 × 1.01 g/mol H) = 78.12 g/mol Moles of condensed benzene = mass / molar mass = 24.7 g (C6H6) / 78.12 g/mol = 0.3161 mol (C6H6)
03

Calculate the initial moles of benzene vapor

Since benzene is condensed, the remaining gaseous mixture in the 100 L container must only be nitrogen gas. Using the Ideal Gas Law, we can determine the initial moles of benzene vapor: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (8.314 J/mol K), and T is the temperature. Rearranging the formula to solve for the number of moles (n): n = PV / RT We know that the volume (V) is 100 L, the temperature (T) is 293.15 K, and R is 8.314 J / (mol K). However, we don't have the pressure yet, but we can find the partial pressure of benzene in the mixture using the condensed moles: Partial pressure of benzene (P_benzene) = Moles of condensed benzene × R × T / V = 0.3161 mol × 8.314 J / (mol K) × 293.15 K / 100 L = 0.9785 atm
04

Calculate the vapor pressure of benzene

Now that we have the partial pressure of benzene, we can calculate the vapor pressure of benzene at 20.0°C. The partial pressure of benzene is equal to its vapor pressure at this temperature: Vapor pressure of benzene = Partial pressure of benzene = 0.9785 atm The vapor pressure of benzene at 20.0°C is 0.9785 atm.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation used to relate different properties of gases. It's expressed as \( PV = nRT \), where \( P \) stands for pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (8.314 J/mol K), and \( T \) represents temperature in Kelvin.

When we apply this law, we can compute any one of these variables if we have information about the others. In our context of nitrogen gas, it helps determine how gases behave under certain conditions.
  • Volume \( (V) \) is usually provided or measured in liters.
  • The temperature \( (T) \) must always be in Kelvin. Convert Celsius to Kelvin by adding 273.15.
  • The equation also requires knowledge about the number of moles \( (n) \) involved, which directly influences pressure \( (P) \).
Understanding the interaction as described by the Ideal Gas Law is critical when studying systems where multiple gaseous components are involved, such as our benzene and nitrogen mixture.
Molar Mass
Molar Mass is a concept that expresses the mass of one mole of a substance. It is calculated by summing the atomic masses of all the atoms in a molecule. For benzene (\( \text{C}_6\text{H}_6 \)), the calculation goes as follows:
  • Carbon (C) has an atomic mass of 12.01 g/mol. Since benzene contains six carbon atoms, the total mass is \(6 \times 12.01 = 72.06 \) g/mol.
  • Hydrogen (H) has an atomic mass of 1.01 g/mol. With six hydrogen atoms, the total mass is \(6 \times 1.01 = 6.06 \) g/mol.
  • Thus, the molar mass of benzene is \( 72.06 \text{ g/mol} + 6.06 \text{ g/mol} = 78.12 \text{ g/mol} \).
This figure aids in converting from mass to moles, which is a necessary step in various calculations, especially those involving chemical equations or gas laws in solution scenarios.
Nitrogen Gas
Nitrogen gas, with the chemical symbol \( \text{N}_2 \), is the most abundant gas in Earth's atmosphere, making up about 78% of the atmosphere. Its properties as an inert gas make it especially useful in various applications where reactive gases could pose a problem.

In experiments, nitrogen is often used as a carrier gas because it doesn’t easily react with other elements or compounds. In the given exercise, it acts as a medium to transport benzene in gas form without chemically interacting with it, allowing benzene to condense elsewhere while nitrogen remains gaseous.

Because nitrogen behaves closely to that of an ideal gas under many conditions, using the Ideal Gas Law with nitrogen often yields accurate predictions in experiments.
Partial Pressure
Partial Pressure refers to the pressure that is exerted by a single type of gas in a mixture of gases. According to Dalton’s Law, the total pressure of a gaseous mixture is the sum of the partial pressures of each individual gas. This is particularly useful when studying gas-liquid systems like our benzene and nitrogen mixture.
  • The Partial Pressure of benzene in the mixture is calculated using its number of moles, the gas constant, and the temperature divided by the volume of the mixture.
  • In our exercise, by knowing the moles of benzene that were condensed (0.3161 mol) and using the ideal conditions, its partial pressure was found to match its vapor pressure.
  • This means that at equilibrium, the pressure exerted by benzene molecules in their gaseous state is equal to the vapor pressure of benzene, verifying the equilibrium state at 20°C.
This understanding is vital when determining how much of a substance will transition between phases or react.

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