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: Problem Identification. A solution contains glucose, a non-volatile, non-electrolyte solute with a molar mass of 180 g mol\(^{-1}\). The task is to determine the solution's freezing point, given information about its boiling point elevation. The boiling point elevation is calculated using the formula: \[ \Delta T_b = K_b \cdot m \] Here, \(\Delta T_b\) denotes the boiling point elevation, \(K_b\) is the ebullioscopic constant (0.512 K kg mol\(^{-1}\)), and \(m\) represents the molality of the solution.

Step 2: Molality Calculation. The solution boils at 100.20°C. The normal boiling point of water is 100°C, leading to a boiling point elevation of: \[ \Delta T_b = 100.20°C - 100°C = 0.20°C \] Applying the boiling point elevation formula: \[ \Delta T_b = K_b \cdot m \] \[ 0.20 = 0.512 \cdot m \] Solving for molality \(m\): \[ m = \frac{0.20}{0.512} = 0.3906 \, \text{mol/kg} \] 

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

Step 4: Freezing Point Determination. The normal freezing point of water is 0°C. Due to the freezing point depression, the freezing point of the solution is: \[ \text{Freezing point} = 0°C - 0.726°C = -0.726°C \] The freezing point of the solution is -0.726°C. \vspace{10pt}