What will be the freezing point of a 0.5 m KCl solution ? The molal freezing point constant of water is `1.86^@C m^(-1)` .

Options

A. `-1.86^@C`
B. `-0.372^@C`
C. `-3.2^@C`
D. `0^@C`

Explanation

To find the freezing point of a 0.5 m KCl solution, we will use the formula for freezing point depression:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \(\Delta T_f\) is the change in freezing point,
- \(i\) is the van 't Hoff factor (the number of particles the solute dissociates into),
- \(K_f\) is the molal freezing point constant,
- \(m\) is the molality of the solution.

### Step 1: Identify the values
- For KCl, the van 't Hoff factor \(i = 2\) (since KCl dissociates into K\(^+\) and Cl\(^-\)).
- The molal freezing point constant \(K_f = 1.86 \, ^\circ C \, m^{-1}\).
- The molality \(m = 0.5 \, m\).

### Step 2: Substitute the values into the formula
\[ \Delta T_f = i \cdot K_f \cdot m \]
\[ \Delta T_f = 2 \cdot 1.86 \, ^\circ C \, m^{-1} \cdot 0.5 \, m \]

### Step 3: Calculate \(\Delta T_f\)
\[ \Delta T_f = 2 \cdot 1.86 \cdot 0.5 = 1.86 \, ^\circ C \]

### Step 4: Determine the freezing point of the solution
The normal freezing point of water is \(0 \, ^\circ C\). The freezing point of the solution will be:
\[ T_f = T_0 - \Delta T_f \]
Where \(T_0 = 0 \, ^\circ C\):
\[ T_f = 0 - 1.86 = -1.86 \, ^\circ C \]

### Final Answer
The freezing point of the 0.5 m KCl solution is \(-1.86 \, ^\circ C\).

---