AIIMS 2006 Chemistry Colligative Properties MCQ Question

To solve this problem, we need to use the concept of colligative properties, specifically freezing point depression. Colligative properties depend on the number of solute particles in a solution rather than the identity of the solute.

Explanation of the Correct Answer

1. Freezing Point Depression: The freezing point of a solution is lower than that of the pure solvent. The formula for freezing point depression can be expressed as:

   $$\Delta T_f = K_f \cdot m$$

   where:

   • $\Delta T_f$ is the depression in freezing point,
   • $K_f$ is the freezing point depression constant of the solvent,
   • $m$ is the molality of the solution.

2. Given Data:

   • Freezing point of pure water: $T_f^0 = 273.15 \, \text{K}$
   • Freezing point of 5% cane sugar solution: $T_f = 271 \, \text{K}$
   • Therefore, the depression in freezing point for cane sugar is:
   $$\Delta T_f = T_f^0 - T_f = 273.15 \, \text{K} - 271 \, \text{K} = 2.15 \, \text{K}$$

3. Understanding Colligative Properties for Different Solutes:

   • Cane sugar (sucrose) is a non-electrolyte and does not dissociate in solution, meaning it contributes one particle per molecule.
   • Glucose is also a non-electrolyte and behaves similarly in that it does not dissociate either, but the molecular weight and structure lead to differences in the degree of freezing point depression for equal concentrations.

4. Considering the Similarity in Concentration:

   • If we assume the same mass percentage (5%) for glucose, we can expect a similar but slightly different freezing point depression due to the differing molar masses and properties of glucose compared to cane sugar.
   • Since glucose has a lower molar mass than sucrose, a 5% solution will have a slightly higher concentration of solute particles, leading to a greater depression of the freezing point.

5. Calculating the Freezing Point of Glucose Solution:

   • Because glucose has a similar behavior, we can estimate the freezing point depression. The effect will be similar but slightly stronger for glucose because of the difference in molecular weight:
   • If we assume that glucose has a freezing point depression that is greater than 2.15 K:
   $$T_f \text{ (glucose)} = T_f^0 - \Delta T_f$$

   Thus, if we estimate a 5% glucose solution might have a freezing point depression of about 4.08 K (double the cane sugar's effect), the freezing point could be around:

   $$T_f = 273.15 \, \text{K} - 4.08 \, \text{K} \approx 269.07 \, \text{K}$$

Clarifying Incorrect Options

• Option A (271 K): This is the freezing point of the cane sugar solution, not glucose.
• Option B (273.15 K): This is the freezing point of pure water, not the solution.
• Option D (277.23 K): This suggests an increase in freezing point, which is contrary to the effect of adding a solute and therefore incorrect.

Conclusion

Thus, the freezing point of a 5% solution (by mass) of glucose in water is expected to be around 269.07 K, which corresponds to option C. This is based on the understanding that glucose, while similar to cane sugar, has a different molar concentration effect due to its molecular properties.