Question

What is the freezing point of a solution containing 1.8 g glucose dissolved in 1000 g of water (Kf of water = 1.86 K kg mol\(^{-1}\) and at mass C=12, H=1, O=16)?

• A. 0.0186°C
• B. -0.0093°C
• C. -0.0186°C
• D. -0.0372°C

Hint:

To calculate the freezing point depression, use the formula \( \Delta T_f = K_f \times m \), where \( m \) is the molality of the solution.

Answer 1

The Correct Option is C

Step 1: Understanding freezing point depression.
The freezing point depression is given by the formula: \[ \Delta T_f = K_f \times m \] Where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the cryoscopic constant, and \( m \) is the molality of the solution.
Step 2: Calculating the molality.
The molar mass of glucose (C6H12O6) is calculated as: \[ M = (6 \times 12) + (12 \times 1) + (6 \times 16) = 180 \, \text{g/mol} \] Molality \( m \) is given by: \[ m = \frac{\text{mol of solute}}{\text{kg of solvent}} = \frac{\frac{1.8}{180}}{1} = 0.01 \, \text{mol/kg} \] Step 3: Applying the formula.
Using \( K_f = 1.86 \, \text{K kg/mol} \) and \( m = 0.01 \, \text{mol/kg} \): \[ \Delta T_f = 1.86 \times 0.01 = 0.0186 \, \text{K} \] Since the solution is freezing, the freezing point will be depressed by 0.0186°C, so the freezing point is: \[ 0°C - 0.0186°C = -0.0186°C \] Step 4: Conclusion.
The correct answer is (C) -0.0186°C, the freezing point of the solution.