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

Step 1: Understanding the problem. We are given that the solution is made with glucose, which is a non-volatile, non-electrolyte solute, and its molar mass is 180 g mol\(^{-1}\). We need to calculate the freezing point of this solution, given the boiling point elevation. The boiling point elevation formula is: \[ \Delta T_b = K_b \cdot m \] Where: \(\Delta T_b\) = Boiling point elevation \(K_b\) = ebullioscopic constant (given as 0.512 K kg mol\(^{-1}\)) \(m\) = molality of the solution

Step 2: Calculate the molality of the solution. The solution’s boiling point is given as 100.20°C. The normal boiling point of water is 100°C, so: \[ \Delta T_b = 100.20°C - 100°C = 0.20°C \] Using the boiling point elevation formula: \[ \Delta T_b = K_b \cdot m \] \[ 0.20 = 0.512 \cdot m \] Solving for molality: \[ m = \frac{0.20}{0.512} = 0.3906 \, \text{mol/kg} \]

Step 3: Calculate the freezing point depression. The freezing point depression formula is: \[ \Delta T_f = K_f \cdot m \] Where: \(\Delta T_f\) = Freezing point depression \(K_f\) = cryoscopic constant (given as 1.86 K kg mol\(^{-1}\)) \(m\) = molality of the solution (calculated as 0.3906 mol/kg) Substituting the values: \[ \Delta T_f = 1.86 \cdot 0.3906 = 0.726 \, \text{°C} \]

Step 4: Calculate the freezing point. The normal freezing point of water is 0°C. Since the solution has a freezing point depression, the freezing point of the solution will be: \[ \text{Freezing point} = 0°C - 0.726°C = -0.726°C \] Thus, the freezing point of the solution is -0.726°C.