Question

A solution of glucose (molar mass = 180 g mol\(^{-1}\)) in water has a boiling point of 100.20°C. Calculate the freezing point of the same solution. Molal constants for water \(K_f\) and \(K_b\) are 1.86 K kg mol\(^{-1}\) and 0.512 K kg mol\(^{-1}\) respectively.

Hint:

Freezing point depression and boiling point elevation are both colligative properties that depend on the molality of the solute and the solvent's cryoscopic and ebullioscopic constants.

Answer 1

1. Determining the Freezing Point of a Glucose Solution:

Provided Information:
- Molar mass of glucose (C₆H₁₂O₆): 180 g/mol
- Solution's boiling point: 100.20°C
- Water's boiling point elevation constant (\(K_b\)): 0.512 K·kg/mol
- Water's freezing point depression constant (\(K_f\)): 1.86 K·kg/mol
- Normal boiling point of water: 100°C
- Normal freezing point of water: 0°C

Step 1: Calculate Boiling Point Elevation:
Boiling point elevation (\(\Delta T_b\)) is the difference between the solution's boiling point and pure water's boiling point:

\(\Delta T_b = 100.20°C - 100°C = 0.20°C\)

Step 2: Calculate Molality using Boiling Point Elevation Formula:
The relationship is \(\Delta T_b = K_b \cdot m\). Solving for molality (\(m\)):

\(m = \frac{\Delta T_b}{K_b} = \frac{0.20}{0.512} = 0.3906 \, \text{mol/kg}\)

Step 3: Calculate Freezing Point Depression:
Using the freezing point depression formula, \(\Delta T_f = K_f \cdot m\):

\(\Delta T_f = 1.86 \cdot 0.3906 = 0.726 \, \text{K}\)

Step 4: Determine the Solution's Freezing Point:
The freezing point depression (\(\Delta T_f\)) is the difference between pure water's freezing point and the solution's freezing point. Therefore, the solution's freezing point is:

Solution freezing point = 0°C - 0.726°C = -0.726°C

Conclusion:
The freezing point of the glucose solution is -0.726°C.