Introduction to Raoult’s Law

Raoult’s Law is a fundamental principle in chemistry that describes how the vapor pressure of a solvent in a solution is affected by the presence of solute. Specifically, it states that the partial vapor pressure of each component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. Understanding how to calculate mole fractions using Raoult’s Law is essential in both laboratory and industrial settings.

Example 1: Calculating the Mole Fraction of Water in a Salt Solution

In a laboratory, a chemist is preparing a saline solution for an experiment. The solution consists of 100 grams of sodium chloride (NaCl) dissolved in 500 grams of water (H₂O). To determine the mole fraction of water in the solution using Raoult’s Law, the chemist first needs to calculate the number of moles of each component.

1. Calculate the number of moles of sodium chloride:

   • Molar mass of NaCl = 58.44 g/mol
   • Moles of NaCl = 100 g / 58.44 g/mol = 1.71 moles

2. Calculate the number of moles of water:

   • Molar mass of H₂O = 18.02 g/mol
   • Moles of H₂O = 500 g / 18.02 g/mol = 27.76 moles

3. Calculate the total number of moles in the solution:

   • Total moles = moles of NaCl + moles of H₂O = 1.71 + 27.76 = 29.47 moles

4. Calculate the mole fraction of water:

   • Mole fraction of H₂O = moles of H₂O / total moles = 27.76 / 29.47 = 0.94

Notes: This example illustrates the use of Raoult’s Law in a simple aqueous solution, demonstrating how to derive mole fractions based on mass and molarity.

Example 2: Determining the Vapor Pressure of a Mixture of Ethanol and Water

An industrial chemist needs to determine the vapor pressure of a mixture containing 200 grams of ethanol (C₂H₅OH) and 800 grams of water at 25°C. Ethanol has a vapor pressure of 0.59 atm at this temperature. The chemist will calculate the mole fraction of ethanol and subsequently the vapor pressure of the solution.

1. Calculate the number of moles of ethanol:

   • Molar mass of ethanol = 46.07 g/mol
   • Moles of ethanol = 200 g / 46.07 g/mol = 4.34 moles

2. Calculate the number of moles of water:

   • Molar mass of water = 18.02 g/mol
   • Moles of water = 800 g / 18.02 g/mol = 44.43 moles

3. Calculate total moles:

   • Total moles = moles of ethanol + moles of water = 4.34 + 44.43 = 48.77 moles

4. Calculate the mole fraction of ethanol:

   • Mole fraction of ethanol = moles of ethanol / total moles = 4.34 / 48.77 = 0.089

5. Calculate the vapor pressure of the solution:

   • Vapor pressure of the solution = (mole fraction of ethanol) × (vapor pressure of pure ethanol)
   • Vapor pressure of the solution = 0.089 × 0.59 atm = 0.0525 atm

Notes: This example showcases the application of Raoult’s Law in determining the vapor pressure in a common mixture, emphasizing the relationship between mole fraction and vapor pressure.

Example 3: Using Raoult’s Law in a Binary Liquid Mixture

In the production of a binary liquid mixture, a technician combines 150 grams of benzene (C₆H₆) and 100 grams of toluene (C₇H₈). The technician aims to find the mole fraction of benzene in the mixture to predict how the properties of the solution will change. The vapor pressure of pure benzene at 25°C is 0.126 atm, while that of pure toluene is 0.028 atm.

1. Calculate the number of moles of benzene:

   • Molar mass of benzene = 78.11 g/mol
   • Moles of benzene = 150 g / 78.11 g/mol = 1.92 moles

2. Calculate the number of moles of toluene:

   • Molar mass of toluene = 92.14 g/mol
   • Moles of toluene = 100 g / 92.14 g/mol = 1.08 moles

3. Calculate total moles:

   • Total moles = moles of benzene + moles of toluene = 1.92 + 1.08 = 3.00 moles

4. Calculate the mole fraction of benzene:

   • Mole fraction of benzene = moles of benzene / total moles = 1.92 / 3.00 = 0.64

5. Calculate the vapor pressure of the solution:

   • Vapor pressure of the solution = (mole fraction of benzene × vapor pressure of pure benzene) + (mole fraction of toluene × vapor pressure of pure toluene)
   • Vapor pressure of the solution = (0.64 × 0.126 atm) + (0.36 × 0.028 atm) = 0.0796 atm + 0.0101 atm = 0.0897 atm

Notes: This example demonstrates how to apply Raoult’s Law in a binary mixture, providing insight into how mole fractions influence the overall vapor pressure of a solution.