Problem 76 - Chemistry: An Atoms-Focused Approach

Question

Arrange the following solutions in order of increasing freezing point depression: a. \(0.10 \mathrm{m} \mathrm{MgCl}_{2}\) in water, \(i=2.7, K_{t}=1.86^{\circ} \mathrm{C} / \mathrm{m}\) b. \(0.20 m\) toluene in diethyl ether, \(i=1.00\) \(K_{f}=1.79^{\circ} \mathrm{C} / \mathrm{m}\) c. \(0.20 \mathrm{m}\) cthylenc glycol in ethanol, \(i=1.00\) \(K_{\mathrm{f}}=1.99^{\circ} \mathrm{C} / \mathrm{m}\)

Step-by-Step Solution

Answer

Question: Arrange the given solutions in the order of increasing freezing point depression: a) 0.10 m MgCl2 in water, b) 0.20 m toluene in diethyl ether, c) 0.20 m ethylene glycol in ethanol. Answer: The order of increasing freezing point depression is b, c, a.

Step 1: Calculate freezing point depression for solution a

First, we will calculate the freezing point depression for solution a, which contains \(0.10 \mathrm{m} \mathrm{MgCl}_{2}\) in water. The given values are \(i=2.7\) and \(K_{t}=1.86^{\circ} \mathrm{C} / \mathrm{m}\). Using the formula, we get: \(\Delta T_{f(a)} = i \times K_{f} \times m = 2.7 \times 1.86 \times 0.10 = 0.5022^{\circ}\mathrm{C}\)

Step 2: Calculate freezing point depression for solution b

Next, we will calculate the freezing point depression for solution b, which contains \(0.20 m\) toluene in diethyl ether. The given values are \(i=1.00\) and \(K_{f}=1.79^{\circ} \mathrm{C} / \mathrm{m}\). Using the formula, we get: \(\Delta T_{f(b)} = i \times K_{f} \times m = 1.00 \times 1.79 \times 0.20 = 0.3580^{\circ}\mathrm{C}\)

Step 3: Calculate freezing point depression for solution c

Finally, we will calculate the freezing point depression for solution c, which contains \(0.20 \mathrm{m}\) ethylene glycol in ethanol. The given values are \(i=1.00\) and \(K_{\mathrm{f}}=1.99^{\circ} \mathrm{C} / \mathrm{m}\). Using the formula, we get: \(\Delta T_{f(c)} = i \times K_{f} \times m = 1.00 \times 1.99 \times 0.20 = 0.3980^{\circ}\mathrm{C}\)

Step 4: Arrange the solutions in order of increasing freezing point depression

Now that we have the freezing point depressions for all three solutions, we can arrange them in increasing order: \(\Delta T_{f(b)} = 0.3580^{\circ} \mathrm{C} < \Delta T_{f(c)} = 0.3980^{\circ} \mathrm{C} < \Delta T_{f(a)} = 0.5022^{\circ} \mathrm{C}\) Thus, the order of increasing freezing point depression is: b, c, a.

Key Concepts

Colligative Properties

Colligative properties are fascinating as they depend solely on the number of solute particles in a solution rather than their identity. These properties include boiling point elevation, vapor pressure lowering, osmotic pressure, and most relevant here, the freezing point depression. Freezing point depression occurs when a solute is added to a solvent, causing the solution to freeze at a lower temperature than the pure solvent.

This phenomenon is crucial in many real-world applications. For example, salting roads in winter lowers the freezing point of water, preventing ice from forming. The degree to which the freezing point is lowered can be calculated using the formula:
\[\Delta T_f = i \times K_f \times m\]
where:

• \( \Delta T_f \) is the freezing point depression,
• \( i \) is the van 't Hoff factor,
• \( K_f \) is the cryoscopic constant specific to the solvent,
• \( m \) is the molality of the solution.

Colligative properties help us understand how solutions behave and allow us to predict how different solutes can affect various solvents.

Van 't Hoff Factor

The van 't Hoff factor \( (i) \) is a key concept when considering colligative properties. It represents the number of particles a compound dissociates into when it dissolves. For non-electrolytes, which do not dissociate in solution, the van 't Hoff factor is typically \( i = 1 \).

Electrolytes, on the other hand, dissociate into ions, leading to a van 't Hoff factor greater than 1. For example, magnesium chloride \( \text{MgCl}_2 \) dissociates into three ions (one \( \text{Mg}^{2+} \) and two \( \text{Cl}^- \)), often giving a theoretical \( i \) value of 3. However, interactions such as ion pairing can reduce this value, resulting in an experimental \( i \) that is slightly less, such as 2.7 in the exercise above.

The van 't Hoff factor is essential for accurately predicting colligative behaviors because it directly affects the calculated value of freezing point depression or any other colligative property.

Molality

Molality is another important concept in understanding colligative properties. It's a concentration unit defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which is based on volume, molality is purely mass-based, making it unaffected by temperature and pressure changes.

This stability makes molality particularly useful when dealing with colligative properties like freezing point depression. When calculating this depression, the formula \( \Delta T_f = i \times K_f \times m \) uses molality \( m \) to ensure that the impact on the solvent's freezing point is precisely calculated, regardless of any volume changes.

By understanding and using molality, we can better predict how a given amount of solute will alter the properties of a solvent, leading to more accurate predictions and adjustments in both laboratory settings and real-world applications.