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.
Molality of the solution is

Options

A. `0.322`
B. `0.222`
C. `0.413`
D. `0.5`

Explanation

To determine the molality of the solution, we can use the freezing point depression formula. The steps are as follows:

### Step 1: Identify the freezing point depression
The freezing point depression (ΔTf) can be calculated using the formula:
\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} \]
Where:
- \(T_f^{\text{pure}} = 273 \, \text{K}\) (freezing point of pure water)
- \(T_f^{\text{solution}} = 272.4 \, \text{K}\) (freezing point of the solution)

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

### Step 2: Use the freezing point depression formula
The freezing point depression is also given by the formula:
\[ \Delta T_f = K_f \cdot m \]
Where:
- \(K_f = 1.86 \, \text{K kg mol}^{-1}\) (freezing point depression constant)
- \(m\) is the molality of the solution.

### Step 3: Rearrange the formula to solve for molality
We can rearrange the formula to solve for molality:
\[ m = \frac{\Delta T_f}{K_f} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ m = \frac{0.6 \, \text{K}}{1.86 \, \text{K kg mol}^{-1}} \]

### Step 5: Calculate the molality
Now, performing the calculation:
\[ m = \frac{0.6}{1.86} \approx 0.322 \, \text{mol kg}^{-1} \]

### Final Answer
The molality of the solution is approximately \(0.322 \, \text{mol kg}^{-1}\).

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