Depression of freezing point of which of the following solutions does represent the cryoscopic constant of water ?

Options

A. 6% by mass of urea in aqueous solution
B. 100 g of aqueous solution containing 18 g of glucose
C. 59 g of aqueous solution containing 9 of glucose
D. 1 M KCI solution in water

Answer

To solve the question regarding the depression of freezing point that represents the cryoscopic constant of water, we will follow these steps:

### Step-by-step Solution:

1. **Understand the Concept**: The depression of freezing point (\( \Delta T_F \)) is given by the formula:
\[ \Delta T_F = i \cdot K_F \cdot M \]
where:
- \( i \) = van't Hoff factor (number of particles the solute breaks into)
- \( K_F \) = cryoscopic constant of the solvent (water in this case)
- \( M \) = molality of the solution

2. **Identify the Condition for Cryoscopic Constant**: We want to find a solution where:
\[ \Delta T_F = K_F \]
This occurs when \( i = 1 \) and \( M = 1 \) molal.

3. **Analyze the Given Solution**: We are given a solution that contains 9 grams of glucose in 59 grams of solution. Since glucose is a non-volatile solute, its van't Hoff factor \( i = 1 \).

4. **Calculate the Weight of the Solvent**:
- Total weight of the solution = 59 grams
- Weight of solute (glucose) = 9 grams
- Therefore, weight of solvent (water) = 59 g - 9 g = 50 grams.

5. **Convert Weight of Solvent to Kilograms**:
\[ \text{Weight of solvent in kg} = \frac{50 \text{ grams}}{1000} = 0.050 \text{ kg} \]

6. **Calculate the Molar Mass of Glucose**: The molar mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol.

7. **Calculate the Molality (M)**:
\[ M = \frac{\text{Weight of solute (in grams)}}{\text{Molar mass of solute (in g/mol)} \times \text{Weight of solvent (in kg)}} \]
Substituting the values:
\[ M = \frac{9 \text{ g}}{180 \text{ g/mol} \times 0.050 \text{ kg}} = \frac{9}{9} = 1 \text{ molal} \]

8. **Conclusion**: Since \( i = 1 \) and \( M = 1 \), we find that:
\[ \Delta T_F = i \cdot K_F \cdot M = 1 \cdot K_F \cdot 1 = K_F \]
Therefore, the depression of freezing point of this solution does represent the cryoscopic constant of water.

### Final Answer:
The correct option is **59 grams of aqueous solution containing 9 grams of glucose**.

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