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.
Depression in freezing point of solution

Options

A. `0.68`
B. `0.43`
C. `0.5989`
D. `0.326`

Explanation

To determine the depression in freezing point of the solution, we can follow these steps:

### Step 1: Identify the freezing points
- The freezing point of pure water is given as 273 K.
- The freezing point of the aqueous solution is given as 272.4 K.

### Step 2: Calculate the depression in freezing point
- The depression in freezing point (ΔTf) can be calculated using the formula:
\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} \]
- Substituting the values:
\[ \Delta T_f = 273 \, \text{K} - 272.4 \, \text{K} \]

### Step 3: Perform the calculation
- Calculate ΔTf:
\[ \Delta T_f = 273 - 272.4 = 0.6 \, \text{K} \]

### Final Answer
- The depression in freezing point of the solution is **0.6 K**.

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