Calculating Vapor Pressure of Solutions

Understanding Vapor Pressure and Raoult’s Law

Vapor pressure is a key concept in chemistry that describes the pressure exerted by a vapor in equilibrium with its liquid or solid phase. Raoult’s Law states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent in the solution. This principle is particularly useful in predicting how the addition of a solute affects the vapor pressure of a solvent. Here are three diverse practical examples of calculating vapor pressure of a solution.

Example 1: Calculating Vapor Pressure of a Saltwater Solution

In this example, we will calculate the vapor pressure of a saltwater solution, which is commonly encountered in ocean studies and environmental science.

Assume we have a solution of 100 grams of salt (NaCl) dissolved in 1,000 grams of water (H₂O). The vapor pressure of pure water at 25°C is 23.76 mmHg.

1. Calculate the moles of NaCl:
   Molar mass of NaCl = 58.44 g/mol
   Moles of NaCl = 100 g / 58.44 g/mol = 1.71 moles

2. Calculate the moles of H₂O:
   Molar mass of H₂O = 18.02 g/mol
   Moles of H₂O = 1,000 g / 18.02 g/mol = 55.51 moles

3. Calculate the mole fraction of H₂O:
   Mole fraction of H₂O = Moles of H₂O / (Moles of H₂O + Moles of NaCl)
   Mole fraction of H₂O = 55.51 / (55.51 + 1.71) = 0.970

4. Calculate the vapor pressure of the solution:
   Vapor pressure of solution = Mole fraction of H₂O × Vapor pressure of pure water
   Vapor pressure of solution = 0.970 × 23.76 mmHg = 23.06 mmHg

Notes: This example illustrates how the presence of a solute (salt) lowers the vapor pressure of the solvent (water) compared to its pure state, which is a common phenomenon in solutions.

Example 2: Determining Vapor Pressure of Ethanol and Water Mixture

This example focuses on a solution where ethanol is mixed with water, commonly used in beverage and chemical industries.

Consider a solution made from 50 grams of ethanol (C₂H₅OH) mixed with 150 grams of water. The vapor pressure of pure ethanol at 25°C is 59 mmHg, and the vapor pressure of pure water at the same temperature is 23.76 mmHg.

1. Calculate the moles of ethanol:
   Molar mass of ethanol = 46.07 g/mol
   Moles of ethanol = 50 g / 46.07 g/mol = 1.09 moles

2. Calculate the moles of water:
   Molar mass of water = 18.02 g/mol
   Moles of water = 150 g / 18.02 g/mol = 8.32 moles

3. Calculate the mole fractions:
   Mole fraction of ethanol = 1.09 / (1.09 + 8.32) = 0.116
   Mole fraction of water = 8.32 / (1.09 + 8.32) = 0.884

4. Calculate the vapor pressures:
   Vapor pressure of ethanol in the mixture = 0.116 × 59 mmHg = 6.84 mmHg
   Vapor pressure of water in the mixture = 0.884 × 23.76 mmHg = 21.02 mmHg

5. Calculate the total vapor pressure:
   Total vapor pressure = Vapor pressure of ethanol + Vapor pressure of water
   Total vapor pressure = 6.84 mmHg + 21.02 mmHg = 27.86 mmHg

Notes: This example demonstrates how to calculate the individual and total vapor pressures of a solution containing two volatile components, highlighting the interactive effects of different solutes.

Example 3: Calculating Vapor Pressure of a Sugar Solution

This example looks at a sugar solution, which is commonly used in food chemistry and culinary applications.

Suppose we dissolve 200 grams of sucrose (C₁₂H₂₂O₁₁) in 1,000 grams of water. The vapor pressure of pure water at 25°C is 23.76 mmHg.

1. Calculate the moles of sucrose:
   Molar mass of sucrose = 342.30 g/mol
   Moles of sucrose = 200 g / 342.30 g/mol = 0.58 moles

2. Calculate the moles of water:
   Molar mass of water = 18.02 g/mol
   Moles of water = 1,000 g / 18.02 g/mol = 55.51 moles

3. Calculate the mole fraction of water:
   Mole fraction of water = 55.51 / (55.51 + 0.58) = 0.994

4. Calculate the vapor pressure of the solution:
   Vapor pressure of solution = 0.994 × 23.76 mmHg = 23.61 mmHg

Notes: In this example, the presence of the non-volatile solute (sucrose) has a minimal effect on the vapor pressure, which is characteristic of non-volatile solutes in dilute solutions. This illustrates the practical application of Raoult’s Law in food science.