Problem 44 - Chemistry: Principles and Reactions

Question

A solution consisting of \(4.50 \mathrm{~g}\) of propylene glycol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{2}\), in \(40.5 \mathrm{~mL}\) of t-butyl alcohol \((d=0.780 \mathrm{~g} / \mathrm{mL})\) freezes at \(8.5^{\circ} \mathrm{C}\). The freezing point of \(\mathrm{t}\) -butyl alcohol is \(25.5^{\circ} \mathrm{C}\). What is its freezing-point constant?

Step-by-Step Solution

Answer

In this problem, we calculated the freezing point constant (Kf) for t-butyl alcohol, given the amounts and properties of solute (propylene glycol) and solvent (t-butyl alcohol). We calculated the freezing point depression (∆Tf) to be 17°C, obtained the mass of the solvent (0.03159 kg), found the molality of the solute (1.875 mol/kg), and used these values to find the freezing point constant (Kf) as approximately 9.07 °C/mol/kg.

Step 1: Calculate freezing point depression (∆Tf)

We need to subtract the freezing point of the solution from the freezing point of the pure solvent: ∆Tf = T_pure - T_solution = 25.5 - 8.5 = 17°C

Step 2: Calculate the mass of solvent

Given the volume of t-butyl alcohol (40.5 mL), which has its density as 0.780 g/mL, we can find the mass of the solvent: Mass_solvent = Volume × Density = 40.5 × 0.780 = 31.59 g

Step 3: Calculate molality

We have been given 4.5 g of propylene glycol, C3H8O2. First, we need to find the molar mass of propylene glycol: Molar mass (C3H8O2) = (3 × 12.01) + (8 × 1.01) + (2 × 16.00) = 76.095 g/mol Now we can find the moles of propylene glycol using the mass: Moles_solute = Mass / Molar_mass = 4.5 / 76.095 ≈ 0.0592 mol Next, convert the mass of the solvent to kg: Mass_solvent (kg) = 31.59 g / 1000 = 0.03159 kg Finally, we can calculate the molality (moles of solute per kg of solvent): molality = Moles_solute / Mass_solvent = 0.0592 / 0.03159 ≈ 1.875 mol/kg

Step 4: Calculate the freezing point constant (Kf)

Knowing the freezing point depression, molality, and the Van't Hoff factor (which is 1 for non-electrolytes), we can easily find Kf: Kf = ∆Tf / (molality × i) = 17 / (1.875 × 1) ≈ 9.07 °C/mol/kg Therefore, the freezing-point constant for t-butyl alcohol is 9.07 °C/mol/kg.

Key Concepts

Colligative Properties

Colligative properties are certain properties of solutions that depend only on the number of dissolved particles in the solution, not on the specific type of particle. This remarkable feature means that whether you dissolve sugar, salt, or any other solute, as long as the amount of moles is the same, the impact on these properties will be identical.

The four main colligative properties are:

• Vapor pressure lowering
• Boiling point elevation
• Freezing point depression
• Osmotic pressure

One key point in understanding colligative properties is the principle that adding a solute to a solvent will alter the physical properties of that solvent. For example, in the case of freezing point depression, the presence of solute particles disrupts the process of solvent solidification, thus lowering the temperature at which the solution freezes compared to the pure solvent.

Molality Calculation

Molality is a measure of concentration used for expressing the amount of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which is affected by changes in volume due to temperature, molality depends solely on the mass, making it temperature-independent.

To calculate molality, you first determine the moles of the solute using its mass and molar mass with the equation:
\[ \text{Moles of solute} = \frac{\text{Mass of solute (g)}}{\text{Molar mass (g/mol)}} \]
Then, you divide the number of moles by the mass of the solvent in kilograms:
\[ \text{Molality (mol/kg)} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} \]
This calculation is crucial for solving problems involving colligative properties, as seen in the provided exercise.

Freezing-Point Constant

The freezing-point constant, also known as the cryoscopic constant, denoted as Kf, is a proportionality factor that relates the freezing point depression of a solution to its molality. Specifically, it quantifies the lowering of the freezing point experienced by a one molal solution of a non-electrolyte solute.

The formula that connects the freezing-point depression (∆Tf), molality (m), and the freezing-point constant is:
\[ \Delta T_f = K_f \times m \times i \]
Here, 'i' represents the Van't Hoff factor, which accounts for the effect of solute particles on the colligative property. The Van't Hoff factor is 1 for non-electrolytes (as they do not dissociate), whereas for electrolytes, it is equal to the number of particles the compound dissociates into.

The exercise you're working on involves finding the freezing-point constant using known values of freezing-point depression and molality. This constant is unique to each solvent and is essential for predicting how a given solvent's freezing point will change with added solute.