Understanding Vapor Pressure Lowering in Solutions

Vapor pressure lowering is a fundamental concept in chemistry that describes how the presence of solute particles in a solvent reduces the solvent’s vapor pressure. This phenomenon is a direct consequence of Raoult’s Law, which states that the vapor pressure of a solvent in a solution is proportional to the mole fraction of the solvent. Below are three practical examples illustrating this concept in diverse contexts.

Example 1: Salt in Water - Boiling Point Elevation

In cooking, adding salt to water is a common practice, but it also serves a scientific purpose. When salt (sodium chloride) is dissolved in water, it dissociates into its constituent ions, Na⁺ and Cl⁻.

In this case, the presence of these solute particles lowers the vapor pressure of the water. Consequently, the boiling point of the solution increases compared to pure water, a phenomenon known as boiling point elevation. This is why salted water takes longer to boil than plain water.

Calculation:

1. Vapor Pressure of Pure Water at 100°C: 760 mmHg.

2. Mole Fraction of Water (assuming 1 mole of salt in 55.5 moles of water):

   • Total moles = 1 (salt) + 55.5 (water) = 56.5 moles.
   • Mole fraction of water = 55.5 / 56.5 = 0.981.

3. Vapor Pressure of Salty Water:

   • Vapor Pressure = 760 mmHg * 0.981 = 745.56 mmHg.

Notes:

The degree of vapor pressure lowering can be calculated further using colligative properties, which reflect the number of solute particles. Adding more salt will continue to lower the vapor pressure further.

Example 2: Sugar in Soft Drinks - Freezing Point Depression

Soft drinks are a popular beverage choice that often contains a significant amount of sugar. When sugar (sucrose) is dissolved in water, it not only sweetens the drink but also affects its freezing point.

The dissolved sugar molecules interfere with the ability of water molecules to form a solid crystalline structure (ice), thus lowering the freezing point of the solution. This phenomenon is particularly relevant in the context of frozen beverages, where a lower freezing point allows the drink to remain slushy instead of solidifying completely.

Calculation:

1. Freezing Point of Pure Water: 0°C.

2. Mole Fraction of Water (assuming 1 mole of sugar in 55.5 moles of water):

   • Total moles = 1 (sugar) + 55.5 (water) = 56.5 moles.
   • Mole fraction of water = 55.5 / 56.5 = 0.981.

3. Freezing Point Depression:

   • Using the formula ΔTf = i * Kf * m, where i = van ‘t Hoff factor, Kf (for water) = 1.86 °C kg/mol, and m is the molality of the solution.
   • Assuming a molality of 1, ΔTf = 1 * 1.86 * 1 = 1.86°C.
   • New freezing point = 0°C - 1.86°C = -1.86°C.

Notes:

The more sugar added, the further the freezing point drops, which is crucial for maintaining the desired texture in frozen drinks.

Example 3: Ethylene Glycol in Antifreeze - Practical Applications

Ethylene glycol is commonly used as an antifreeze agent in vehicles. When mixed with water, it significantly lowers the vapor pressure of the solution, which is essential for preventing the coolant from evaporating at high temperatures.

This property not only allows the coolant to remain in the liquid state under normal operating conditions but also raises the boiling point of the solution, making it effective for use in engines.

Calculation:

1. Vapor Pressure of Pure Water at 100°C: 760 mmHg.

2. Mole Fraction of Water in 50% Ethylene Glycol Solution:

   • Assuming 1 mole of ethylene glycol and 55.5 moles of water.
   • Total moles = 1 + 55.5 = 56.5 moles.
   • Mole fraction of water = 55.5 / 56.5 = 0.981.

3. Vapor Pressure of Ethylene Glycol Solution:

   • Vapor Pressure = 760 mmHg * 0.981 = 745.56 mmHg.

Notes:

The composition of the antifreeze solution can vary, and it is important to maintain the correct ratios to ensure effective performance across a range of temperatures.