Question

A 5% solution of cane sugar (342 g/mol) in water has a freezing point of 271 K. Find the freezing point of a 5% glucose (180 g/mol) solution. [Water $T_f = 2715$ K]

• A. 271 K
• B. 269 K
• C. 259 K
• D. 273 K

Hint:

Lower molar mass means more solute particles for the same mass of solute, producing larger colligative effect and greater freezing point depression.

Answer 1

The Correct Option is B

Understanding the Concept: Depression in freezing point is given by: \[ \Delta T_f = K_f m \] For solutions having same mass percentage: \[ \Delta T_f \propto \frac{1}{M} \] where:
$M$ = molar mass
Lower molar mass means larger number of solute particles
Greater number of particles causes larger depression in freezing point

Step 1: Calculate Depression in Freezing Point for Cane Sugar} Freezing point of pure water: \[ 273.15\ K \] Freezing point of cane sugar solution: \[ 271\ K \] Therefore: \[ \Delta T_{f1} = 273.15 - 271 \] \[ \Delta T_{f1} = 2.15\ K \]
Step 2: Apply Relation Between Depression and Molar Mass} For same mass percentage: \[ \frac{\Delta T_{f2}}{\Delta T_{f1}} = \frac{M_1}{M_2} \] Substituting values: \[ \frac{\Delta T_{f2}}{2.15} = \frac{342}{180} \] \[ \frac{\Delta T_{f2}}{2.15} = 1.9 \] Therefore: \[ \Delta T_{f2} = 1.9 \times 2.15 \] \[ \Delta T_{f2} \approx 4.08\ K \]
Step 3: Calculate New Freezing Point} \[ T_f = 273.15 - 4.08 \] \[ T_f = 269.07\ K \] Approximately: \[ 269\ K \] Hence, the correct answer is: \[ \boxed{(B)\ 269\ K} \]