Examples of Raoult's Law in Ideal Solutions

Understanding 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 molecules. According to this law, the vapor pressure of an ideal solution is directly proportional to the mole fraction of the solvent. This concept is crucial in various applications, including distillation, solution chemistry, and understanding colligative properties. Below are three diverse examples that illustrate the application of Raoult’s Law in ideal solutions.

Example 1: Calculating Vapor Pressure of a Sugar Solution

In the food industry, it’s common to dissolve sugar in water to create sweetened beverages. When sugar is added to water, it alters the vapor pressure of the solution. By applying Raoult’s Law, we can quantify this change.

In a solution of 100 g of sugar (C₁₂H₂₂O₁₁) dissolved in 900 g of water (H₂O), we first need to calculate the moles of each component:

• Molar mass of water = 18.02 g/mol
• Moles of water = 900 g / 18.02 g/mol = 49.94 moles
• Molar mass of sugar = 342.30 g/mol
• Moles of sugar = 100 g / 342.30 g/mol = 0.292 moles

Now, we can find the mole fraction of water (X₁):

• Total moles = 49.94 + 0.292 = 50.232
• Mole fraction of water (X₁) = 49.94 / 50.232 = 0.993

Assuming the vapor pressure of pure water (P₀) at room temperature is 23.8 mmHg, we apply Raoult’s Law:

• P_solution = X₁ * P₀ = 0.993 * 23.8 mmHg = 23.63 mmHg

Notes: This example illustrates how adding a solute (sugar) decreases the vapor pressure of the solvent (water), which is essential for understanding the sweetening process in beverages.

Example 2: Distillation of Ethanol-Water Mixtures

Distillation is a common method for separating components of a mixture based on their differing vapor pressures. In an ethanol-water mixture, we can employ Raoult’s Law to understand how the two components behave during distillation.

Consider a mixture with 60% ethanol and 40% water by mole. To find the vapor pressures:

• Molar mass of ethanol (C₂H₅OH) = 46.07 g/mol, with a vapor pressure at 25°C (P₀_ethanol) = 78.29 mmHg.
• Molar mass of water = 18.02 g/mol, with a vapor pressure at 25°C (P₀_water) = 23.76 mmHg.

Calculating the mole fractions:

• Moles of ethanol in 1 L of the mixture (assuming density = 0.789 g/mL): 60% of 789 g = 473.4 g => Moles of ethanol = 473.4 g / 46.07 g/mol = 10.27 moles.
• Moles of water (40%): 40% of 1000 g = 400 g => Moles of water = 400 g / 18.02 g/mol = 22.18 moles.

Total moles = 10.27 + 22.18 = 32.45.

• Mole fraction of ethanol (X_ethanol) = 10.27 / 32.45 = 0.316
• Mole fraction of water (X_water) = 22.18 / 32.45 = 0.684

Using Raoult’s Law, we find the partial vapor pressures:

• P_ethanol = X_ethanol * P₀_ethanol = 0.316 * 78.29 mmHg = 24.77 mmHg
• P_water = X_water * P₀_water = 0.684 * 23.76 mmHg = 16.25 mmHg

The total vapor pressure above the mixture (P_total) = P_ethanol + P_water = 24.77 + 16.25 = 41.02 mmHg.

Notes: This example demonstrates how Raoult’s Law aids in predicting the behavior of components in a distillation process, essential for industries like beverage production and chemical manufacturing.

Example 3: Boiling Point Elevation in Saltwater Solutions

In maritime applications, understanding the effects of dissolved salts in water is crucial for predicting boiling point changes. When table salt (NaCl) is added to water, it affects the vapor pressure and, consequently, the boiling point.

Assuming we have a solution containing 58.44 g of NaCl in 1 kg (1000 g) of water:

• Molar mass of NaCl = 58.44 g/mol, thus moles of NaCl = 58.44 g / 58.44 g/mol = 1 mole.
• Moles of water = 1000 g / 18.02 g/mol = 55.51 moles.

The total number of moles in the solution = 1 (NaCl) + 55.51 (water) = 56.51.

Calculating mole fractions:

• Mole fraction of water (X_water) = 55.51 / 56.51 = 0.981

The vapor pressure of pure water at boiling point (P₀_water) = 760 mmHg. Using Raoult’s Law:

• P_solution = X_water * P₀_water = 0.981 * 760 mmHg = 745.56 mmHg.

The decrease in vapor pressure leads to an increase in boiling point. The boiling point elevation can be calculated using the formula:

• ΔT_b = K_b * m, where K_b for water = 0.512 °C kg/mol and m is the molality of the solution. In this case, molality (m) = moles of solute / kg of solvent = 1 mole / 1 kg = 1 mol/kg.
• ΔT_b = 0.512 °C kg/mol * 1 mol/kg = 0.512 °C.

Thus, the new boiling point of the solution = 100 °C + 0.512 °C = 100.512 °C.

Notes: This application is particularly significant in marine engineering and environmental science, where salinity levels can impact boiling points and other physical properties of seawater.