Answer: To calculate the freezing point depression, we can use the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( \Delta T_f \) is the freezing point depression,
- \( i \) is the van't Hoff factor,
- \( K_f \) is the cryoscopic constant (which is 1.86 °C kg/mol for water),
- \( m \) is the molality of the solute.
First, let's calculate the molality of the solute (KCl).
Molar mass of KCl (potassium chloride) = 39.10 g/mol (for potassium) + 35.45 g/mol (for chlorine) = 74.55 g/mol.
Number of moles of KCl = Mass of KCl / Molar mass of KCl
\[ n_{\text{KCl}} = \frac{2 \, \text{g}}{74.55 \, \text{g/mol}} = 0.0268 \, \text{mol} \]
Next, let's calculate the molality of the solute:
\[ m = \frac{n_{\text{KCl}}}{\text{mass of solvent in kg}} \]
The mass of the solvent (water) is 25 g.
\[ m = \frac{0.0268 \, \text{mol}}{0.025 \, \text{kg}} = 1.072 \, \text{mol/kg} \]
Now, we can calculate the freezing point depression:
\[ \Delta T_f = i \cdot K_f \cdot m \]
For KCl, the van't Hoff factor (i) is 2, as it dissociates into two ions in solution (K⁺ and Cl⁻).
\[ \Delta T_f = (2) \cdot (1.86 \, \text{°C kg/mol}) \cdot (1.072 \, \text{mol/kg}) \]
\[ \Delta T_f = 3.97 \, \text{°C} \]
So, the freezing point depression caused by dissolving 2 g of KCl in 25 g of water is approximately 3.97 °C.