An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given `K_(f)=1.86 K kg "mol"^(-1)`,`K_(b)=0.512 K kg "mol"^(-1)` and vapour pressure of water at 298 K = 23.756 mm Hg. Determine the following.
Boiling point of the solution is

Options

A. `300.73 K`
B. `373.165 K`
C. `400 K`
D. `273.15 K`

Explanation

To determine the boiling point of the solution, we will follow these steps:

### Step 1: Calculate the depression in freezing point (ΔTf)
The depression in freezing point can be calculated using the formula:

\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} \]

Where:
- \( T_f^{\text{pure}} \) = 273 K (freezing point of pure water)
- \( T_f^{\text{solution}} \) = 272.4 K (freezing point of the solution)

Calculating ΔTf:

\[ \Delta T_f = 273 \, \text{K} - 272.4 \, \text{K} = 0.6 \, \text{K} \]

### Step 2: Calculate the molality (m) of the solution
Using the formula for freezing point depression:

\[ \Delta T_f = K_f \times m \]

Where:
- \( K_f = 1.86 \, \text{K kg mol}^{-1} \)

Rearranging the formula to find molality (m):

\[ m = \frac{\Delta T_f}{K_f} = \frac{0.6 \, \text{K}}{1.86 \, \text{K kg mol}^{-1}} \approx 0.3226 \, \text{mol kg}^{-1} \]

### Step 3: Calculate the elevation in boiling point (ΔTb)
Using the formula for boiling point elevation:

\[ \Delta T_b = K_b \times m \]

Where:
- \( K_b = 0.512 \, \text{K kg mol}^{-1} \)

Substituting the values:

\[ \Delta T_b = 0.512 \, \text{K kg mol}^{-1} \times 0.3226 \, \text{mol kg}^{-1} \approx 0.165 \, \text{K} \]

### Step 4: Calculate the boiling point of the solution (Tb)
The boiling point of the solution can be calculated using:

\[ T_b = T_b^{\text{pure}} + \Delta T_b \]

Where:
- \( T_b^{\text{pure}} = 373 \, \text{K} \) (boiling point of pure water)

Calculating Tb:

\[ T_b = 373 \, \text{K} + 0.165 \, \text{K} \approx 373.165 \, \text{K} \]

### Final Answer
The boiling point of the solution is approximately **373.165 K**.

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