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The freezing point depression is a fascinating phenomenon that occurs when a solute is dissolved in a solvent. In this case, when HCl is added to water, the result is a lowering of the freezing point of the solution compared to pure water. This effect is due to the presence of solute particles interfering with the formation of the solid structure of ice. Understanding this concept requires knowing that pure water freezes at 0°C. When you dissolve 9.12 g of HCl in 190 g of water, the freezing point of the solution drops to -4.65°C. This change in freezing point is quantified as the freezing point depression (\(\Delta T\)), which is calculated by subtracting the freezing point of the solution from the freezing point of the pure solvent. For this example:

• Freezing point of pure water: 0°C
• Freezing point of solution: -4.65°C
• \(\Delta T = 0°C - (-4.65°C) = 4.65°C\).

This principle not only explains why roads are salted in winter to prevent ice formation but also helps understand the colligative properties in chemistry.

Determining the molality of a solution is a key step in calculating the freezing point depression. Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent. It provides a measure of concentration that is not affected by changes in temperature or pressure, making it particularly useful for studying colligative properties. In the given example, to find the molality of the HCl solution:

• First, convert the mass of HCl from grams to moles. The formula is: \[\text{moles of HCl} = \frac{9.12 \text{ g}}{36.46 \text{ g/mol}} \approx 0.250 \text{ moles}\]
• Next, note that the mass of water, the solvent, is 190 g or 0.19 kg.
• Then, molality is calculated as: \[m = \frac{0.250 \text{ moles}}{0.19 \text{ kg}} \approx 1.316 \text{ mol/kg}\]

This calculation tells us the concentration of HCl in the solution and is essential for determining how much the freezing point is depressed.

Hydrochloric acid (HCl) is a strong acid with distinct properties that affect how it behaves in solution. In water, HCl completely dissociates into hydrogen ions (H\(^+\)) and chloride ions (Cl\(^-\)). This complete dissociation is a defining feature of strong acids. The van't Hoff factor (\(i\)) reflects the number of particles a solute dissociates into in solution and is crucial for accurately predicting colligative properties like freezing point depression. For an ideal scenario, the van't Hoff factor for HCl would be 2, indicating complete dissociation into two ions. In this exercise, after calculating the changes in freezing point due to HCl's presence, the van't Hoff factor is determined:

• Using the formula for freezing point depression: \[\Delta T = i \cdot K_f \cdot m\]
• Where \(\Delta T = 4.65°C\), \(K_f = 1.86°C\,\text{kg/mol}\), and \(m = 1.316\,\text{mol/kg}\).
• The van't Hoff factor is: \[i = \frac{4.65}{1.86 \times 1.316} \approx 1.90\]

This value slightly deviates from the ideal 2, which is common in real-world applications due to ion pairing or other factors affecting dissociation.