Calculate depression of freezing point for 0.56 molal aq. Solution of KCl.
(Given : `K_f(H_(2)O) = 1.8 kg mol^(-1)`).

Answer

To calculate the depression of the freezing point for a 0.56 molal aqueous solution of KCl, we will use the formula for freezing point depression:

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

Where:
- \(\Delta T_f\) = depression in freezing point (in Kelvin)
- \(K_f\) = molal freezing point depression constant of the solvent (for water, \(K_f = 1.8 \, \text{K kg mol}^{-1}\))
- \(m\) = molality of the solution (in mol/kg)

### Step 1: Identify the values
We have:
- \(K_f = 1.8 \, \text{K kg mol}^{-1}\)
- \(m = 0.56 \, \text{molal}\)

### Step 2: Substitute the values into the formula
Now, we substitute the values into the formula:

\[ \Delta T_f = 1.8 \, \text{K kg mol}^{-1} \times 0.56 \, \text{mol/kg} \]

### Step 3: Perform the multiplication
Now we calculate:

\[ \Delta T_f = 1.8 \times 0.56 = 1.008 \, \text{K} \]

### Step 4: Round the result
Since we are typically interested in the depression of freezing point to one decimal place, we can round our result:

\[ \Delta T_f \approx 1.0 \, \text{K} \]

### Final Answer
The depression of the freezing point for the 0.56 molal aqueous solution of KCl is approximately **1.0 K**.

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