If the boiling point of an aqueous solution containing a non-volatile solute is `100.1^(@)C`. What is its freezing point? Given latent heat of fusion and vapourization of water `80 cal g^(-1)` and `540 cal g^(-1)`, respectively.

Options

A. `0.361^(@)C`
B. `-0.361^(@)C`
C. `-3.61^(@)C`
D. None of these

Answer

To solve the problem of finding the freezing point of an aqueous solution containing a non-volatile solute, we can follow these steps:

### Step 1: Understand the Given Information
- Boiling point of the solution, \( T_b = 100.1^\circ C \)
- Boiling point of pure water, \( T_{b0} = 100^\circ C \)
- Latent heat of fusion of water, \( L_f = 80 \, \text{cal/g} \)
- Latent heat of vaporization of water, \( L_v = 540 \, \text{cal/g} \)

### Step 2: Calculate the Elevation in Boiling Point
The elevation in boiling point (\( \Delta T_b \)) is given by:
\[ \Delta T_b = T_b - T_{b0} = 100.1^\circ C - 100^\circ C = 0.1^\circ C \]

### Step 3: Use the Relationship Between Elevation in Boiling Point and Depression in Freezing Point
The relationship between the elevation in boiling point and the depression in freezing point is given by:
\[ \frac{\Delta T_b}{\Delta T_f} = \frac{K_b}{K_f} \]
Where:
- \( K_b \) is the molal boiling point elevation constant.
- \( K_f \) is the molal freezing point depression constant.

### Step 4: Calculate \( K_b \) and \( K_f \)
Using the formulas for \( K_b \) and \( K_f \):
\[ K_b = \frac{R T_b^2}{1000 L_v} \]
\[ K_f = \frac{R T_f^2}{1000 L_f} \]
Where \( R \) is the gas constant (approximately \( 1.987 \, \text{cal/(mol K)} \)).

### Step 5: Substitute Values to Find the Ratio
We can substitute the values into the ratio:
\[ \frac{\Delta T_b}{\Delta T_f} = \frac{\frac{R T_b^2}{1000 L_v}}{\frac{R T_f^2}{1000 L_f}} = \frac{T_b^2 L_f}{T_f^2 L_v} \]

### Step 6: Rearrange to Solve for \( \Delta T_f \)
Rearranging gives:
\[ \Delta T_f = \Delta T_b \cdot \frac{T_f^2 L_v}{T_b^2 L_f} \]

### Step 7: Substitute Known Values
- Convert temperatures to Kelvin:
- \( T_b = 373 \, \text{K} \)
- \( T_f = 273 \, \text{K} \)
- Substitute into the equation:
\[ \Delta T_f = 0.1 \cdot \frac{(273)^2 \cdot 540}{(373)^2 \cdot 80} \]

### Step 8: Calculate \( \Delta T_f \)
Calculating the right-hand side:
\[ \Delta T_f = 0.1 \cdot \frac{74529 \cdot 540}{139129 \cdot 80} \]
\[ = 0.1 \cdot \frac{40245660}{11130320} \approx 0.361 \]

### Step 9: Find the Freezing Point of the Solution
The freezing point of the solution (\( T_f \)) is given by:
\[ T_f = T_{f0} - \Delta T_f = 0 - 0.361 = -0.361^\circ C \]

### Final Answer
The freezing point of the solution is:
\[ T_f \approx -0.361^\circ C \]

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