To calculate the freezing point of a solution, we will use the freezing point depression formula:

$\Delta T_f = i \cdot K_f \cdot m$

Where:

• $\Delta T_f$ is the freezing point depression (the amount the freezing point decreases).
• $i$ is the van 't Hoff factor (for HCl, which dissociates fully into H⁺ and Cl⁻, $i = 2$).
• $K_f$ is the cryoscopic constant for water (which is 1.86°C·kg/mol).
• $m$ is the molality of the solution.

Step-by-step solution:

1. Calculate moles of HCl:

   • Molar mass of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol.
   • Moles of HCl = $\frac{23 \, \text{g}}{36.5 \, \text{g/mol}} \approx 0.63 \, \text{mol}$.

2. Convert 3500 mL of water to kilograms:

   • Water has a density of approximately 1 g/mL.
   • 3500 mL = 3500 g = 3.5 kg.

3. Calculate molality (m):

   • Molality = $\frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{0.63 \, \text{mol}}{3.5 \, \text{kg}} \approx 0.18 \, \text{mol/kg}$.

4. Calculate the freezing point depression: \Delta T_f = 2 \cdot 1.86 \, ^\circ\text{C·kg/mol} \cdot 0.18 \, \text{mol/kg} \approx 0.67^\circ \text{C}

5. Determine the new freezing point:

   • The normal freezing point of water is 0°C.
   • New freezing point = $0^\circ \text{C} - 0.67^\circ \text{C} = -0.67^\circ \text{C}$.

Conclusion:

The freezing point of the solution is approximately -0.67°C.

Would you like more details, or have any further questions?

Here are 5 related questions to expand your understanding:

1. How does the van 't Hoff factor change for different solutes?
2. What is the cryoscopic constant for other solvents besides water?
3. How does molality differ from molarity in solution calculations?
4. How does the concentration of a solute affect the freezing point of a solution?
5. Can freezing point depression be used to calculate the molar mass of an unknown solute?

Tip: When working with colligative properties, it's important to account for the dissociation of the solute into ions, as it affects the van 't Hoff factor.