A 5% solution (w/W) of cane sugar (molar mass = 342 g `mol^(-1)` ) has freezing point 271 K. What will be the freezing point of 5% glucose (molar mass = 18 g `mol^(-1)`) in water if freezing point of pure water is 273.15 K ?

Options

A. 273.07 K
B. 269.07 K
C. 273.15 K
D. 260.09 K

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the depression in freezing point for cane sugar solution.

1. **Identify the initial freezing point of pure water**:
- Freezing point of pure water = 273.15 K
- Freezing point of cane sugar solution = 271 K

2. **Calculate the depression in freezing point (ΔTf)**:
\[ \Delta T_f = T_f(\text{pure solvent}) - T_f(\text{solution}) = 273.15 \, \text{K} - 271 \, \text{K} = 2.15 \, \text{K} \]

### Step 2: Calculate the molality of the cane sugar solution.

1. **Determine the mass of cane sugar in the solution**:
- A 5% (w/w) solution means 5 g of cane sugar in 100 g of solution.

2. **Calculate the number of moles of cane sugar**:
\[ \text{Number of moles of cane sugar} = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \, \text{g}}{342 \, \text{g/mol}} \approx 0.0146 \, \text{mol} \]

3. **Calculate the mass of the solvent (water)**:
- Mass of solvent = Total mass of solution - Mass of solute = 100 g - 5 g = 95 g = 0.095 kg

4. **Calculate the molality (m)**:
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0146 \, \text{mol}}{0.095 \, \text{kg}} \approx 0.1547 \, \text{mol/kg} \]

### Step 3: Calculate the cryoscopic constant (Kf).

1. **Use the formula for depression in freezing point**:
\[ \Delta T_f = K_f \cdot m \]
Rearranging gives:
\[ K_f = \frac{\Delta T_f}{m} = \frac{2.15 \, \text{K}}{0.1547 \, \text{mol/kg}} \approx 13.91 \, \text{K kg/mol} \]

### Step 4: Calculate the freezing point of the glucose solution.

1. **Determine the mass of glucose in the solution**:
- A 5% (w/w) glucose solution means 5 g of glucose in 100 g of solution.

2. **Calculate the number of moles of glucose**:
\[ \text{Number of moles of glucose} = \frac{5 \, \text{g}}{180 \, \text{g/mol}} \approx 0.0278 \, \text{mol} \]

3. **Calculate the mass of the solvent (water)**:
- Mass of solvent = 100 g - 5 g = 95 g = 0.095 kg

4. **Calculate the molality (m)**:
\[ m = \frac{0.0278 \, \text{mol}}{0.095 \, \text{kg}} \approx 0.2926 \, \text{mol/kg} \]

### Step 5: Calculate the depression in freezing point for glucose solution.

1. **Use the cryoscopic constant (Kf)**:
\[ \Delta T_f = K_f \cdot m = 13.91 \, \text{K kg/mol} \cdot 0.2926 \, \text{mol/kg} \approx 4.07 \, \text{K} \]

### Step 6: Calculate the final freezing point of the glucose solution.

1. **Calculate the final freezing point**:
\[ T_f(\text{glucose solution}) = T_f(\text{pure solvent}) - \Delta T_f = 273.15 \, \text{K} - 4.07 \, \text{K} \approx 269.08 \, \text{K} \]

### Final Answer:
The freezing point of the 5% glucose solution is approximately **269.08 K**.

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