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