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. Calculating the Freezing Point of the Glucose Solution:

Given Data:
- Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol
- Boiling point of the solution = 100.20°C
- Boiling point elevation constant for water, \(K_b\) = 0.512 K·kg/mol
- Freezing point depression constant for water, \(K_f\) = 1.86 K·kg/mol
- Normal boiling point of water = 100°C
- Normal freezing point of water = 0°C

Step 1: Calculate the Boiling Point Elevation:
The boiling point elevation (\(\Delta T_b\)) is the difference between the boiling point of the solution and the normal boiling point of water:

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

Step 2: Use the Boiling Point Elevation Formula:
The formula for boiling point elevation is given by:

\(\Delta T_b = K_b \cdot m\)

where \(m\) is the molality of the solution. We can rearrange this to solve for \(m\) (molality):

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

Step 3: Calculate the Freezing Point Depression:
Now, we use the formula for freezing point depression (\(\Delta T_f\)):

\(\Delta T_f = K_f \cdot m\)

Substitute the value of \(m\) and \(K_f\):

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

Step 4: Calculate the Freezing Point of the Solution:
The freezing point depression (\(\Delta T_f\)) is the difference between the normal freezing point of water and the freezing point of the solution:
Freezing point of the solution = 0°C - 0.726°C = -0.726°C

Final Answer:
The freezing point of the solution is -0.726°C.