Understanding Boiling Point Elevation

Boiling point elevation is a colligative property that describes how the boiling point of a solvent increases when a non-volatile solute is added. This phenomenon is significant in various applications, including cooking, chemistry, and even industrial processes. The formula to calculate boiling point elevation is:

[ \Delta T_b = i imes K_b imes m ]

Where:

• (\Delta T_b) = change in boiling point
• (i) = van ‘t Hoff factor (number of particles the solute breaks into)
• (K_b) = ebullioscopic constant of the solvent (°C/m)
• (m) = molality of the solution (moles of solute per kg of solvent)

Example 1: Salt in Water

In cooking, adding salt to water increases the boiling point, allowing food to cook faster. This example demonstrates how to calculate the boiling point elevation of water when salt (sodium chloride, NaCl) is added.

1. Context: You are boiling water for pasta and decide to add 0.5 kg of table salt (NaCl) to 10 kg of water.

2. Calculation:

   • Van ‘t Hoff Factor (i): NaCl dissociates into two ions (Na⁺ and Cl⁻), so (i = 2).
   • Ebullioscopic Constant (K_b): For water, (K_b = 0.512 °C/m).
   • Molality (m):
     • Moles of NaCl = (\frac{0.5 kg}{58.44 g/mol} = 8.55 mol)
     • Molality = (\frac{8.55 mol}{10 kg} = 0.855 m)
   • Boiling Point Elevation (ΔT_b):
     [ \Delta T_b = 2 \times 0.512 °C/m \times 0.855 m = 0.875 °C ]

Thus, the boiling point of the water increases by approximately 0.875 °C, resulting in a new boiling point of about 100.875 °C.

3. Notes: This is a practical example of boiling point elevation in culinary applications. The actual effect may vary based on additional factors like atmospheric pressure.

Example 2: Antifreeze in Automotive Applications

Antifreeze is crucial in preventing engine coolant from freezing in cold temperatures. This example illustrates how to calculate the boiling point elevation when ethylene glycol is added to water in a car’s cooling system.

1. Context: A vehicle’s cooling system contains 15 liters of water, and you add 2 liters of ethylene glycol (C₂H₆O₂).

2. Calculation:

   • Van ‘t Hoff Factor (i): Ethylene glycol does not dissociate, so (i = 1).
   • Ebullioscopic Constant (K_b): For water, (K_b = 0.512 °C/m).
   • Molality (m):
     • Molar Mass of Ethylene Glycol = 62.07 g/mol
     • Moles of Ethylene Glycol = (\frac{2000 g}{62.07 g/mol} = 32.24 mol)
     • Total mass of water = 15 kg (15 liters)
     • Molality = (\frac{32.24 mol}{15 kg} = 2.15 m)
   • Boiling Point Elevation (ΔT_b):
     [ \Delta T_b = 1 \times 0.512 °C/m \times 2.15 m = 1.10 °C ]

Consequently, the boiling point of the coolant increases by approximately 1.10 °C, enhancing engine performance in high temperatures.

3. Notes: This situation illustrates the importance of boiling point elevation in automotive engineering, particularly in cold climates.

Example 3: Freezing Point Depression in Ice Cream Making

While boiling point elevation is often discussed, freezing point depression is another related concept. This example describes how sugar affects the freezing point of a mixture when making ice cream.

1. Context: In preparing ice cream, you combine 1 kg of cream with 0.2 kg of sugar (sucrose, C₁₂H₂₂O₁₁).

2. Calculation:

   • Van ‘t Hoff Factor (i): Sucrose does not dissociate, so (i = 1).
   • Cryoscopic Constant (K_f): For water, (K_f = 1.86 °C/m).
   • Molality (m):
     • Molar Mass of Sucrose = 342.30 g/mol
     • Moles of Sucrose = (\frac{200 g}{342.30 g/mol} = 0.58 mol)
     • Total mass of cream = 1 kg
     • Molality = (\frac{0.58 mol}{1 kg} = 0.58 m)
   • Freezing Point Depression (ΔT_f):
     [ \Delta T_f = 1 \times 1.86 °C/m \times 0.58 m = 1.08 °C ]

Thus, the freezing point of the ice cream mixture decreases by approximately 1.08 °C, which helps achieve a creamy texture.

3. Notes: This example highlights the practical application of colligative properties in food science, particularly in dessert preparation.