Question

Liquid naphthalene normally freezes at \(80.2^{\circ} \mathrm{C}\). When 1 mole of solute is dissolved in \(1000 \mathrm{~g}\) of naphthalene, the freezing point of the solution is \(73.2^{\circ} \mathrm{C}\). When \(6.0 \mathrm{~g}\) of sulfur is dissolved in \(250 \mathrm{~g}\) of naphthalene, the freezing point is \(79.5^{\circ} \mathrm{C}\). What is the molecular weight of sulfur?

Short answer

The molecular weight of sulfur is \(240 \thinspace g/mol\).

Step-by-step solution

Calculate the molality of the solution with 1 mole of solute.

Since there is one mole of solute in 1000 grams of naphthalene, the molality is \(1 \frac{mol}{1000 \cdot 10^{-3} kg} = 1 \frac{mol}{1 kg}\) (where we convert grams to kilograms).

Calculate the change in freezing point for the solution with 1 mole of solute.

Since the freezing point of pure naphthalene is 80.2°C and that of the solution is 73.2°C, ΔT = 80.2 - 73.2 = 7°C.

Determine the value of the freezing point depression constant (Kf).

Using the freezing point depression equation, ΔT = Kf * molality * i, we can now calculate Kf. Since the change in freezing point (ΔT) is 7°C and the molality is 1, we can simplify the equation: \(7 = Kf * 1 * i\), where i is the van't Hoff factor. For non-electrolytes, such as naphthalene, i = 1. Thus, the freezing point depression constant, Kf, is equal to 7.

Calculate the change in freezing point for the solution with 6g of sulfur.

The freezing point of the solution with sulfur is given as 79.5°C, and the freezing point of pure naphthalene is 80.2°C, so ΔT = 80.2 - 79.5 = 0.7°C.

Calculate the molality of the solution with 6g of sulfur.

We know that ΔT = Kf * molality * i, and we have values for ΔT (0.7°C) and Kf (7). Thus, we calculate the molality as follows: \[\frac{0.7}{7} = molality * 1\], which implies that the molality of the sulfur solution is 0.1 mol/kg.

Calculate the number of moles of sulfur in the solution.

Since we have the molality and the mass of naphthalene in the solution, we can find the number of moles of sulfur in the solution: \[0.1\frac{mol}{kg} = \frac{moles \thinspace of \thinspace sulfur}{0.25kg}\] This equation gives us: moles of sulfur = 0.1 * 0.25 = 0.025 mol

Calculate the molecular weight of sulfur.

Now that we have the number of moles of sulfur (0.025 mol) and the mass of sulfur (6g), we can find the molecular weight of sulfur using the equation: \[molecular \thinspace weight = \frac{mass}{moles}\] Thus, the molecular weight of sulfur = \(\frac{6}{0.025} = 240 \thinspace g/mol\).

Key concepts

Freezing Point Depression

Freezing point depression is a colligative property, which means it depends on the number of solute particles in a solvent, not their identity. When a solute is dissolved in a solvent, the freezing point of the resulting solution is lower than that of the pure solvent. This happens because the solute particles disrupt the solvent's ability to form a solid structure, thus requiring a lower temperature to freeze.

In our exercise, the pure naphthalene freezes at 80.2°C, but when a solute is added, the freezing point drops. For example, when 1 mole of solute is added to 1000 grams of naphthalene, the freezing point drops to 73.2°C. This change in temperature is directly related to the amount of solute and serves as a foundation for calculating properties like molality and the freezing point depression constant.

Molality

Molality is a measure of the concentration of a solution and is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, molality is not affected by changes in temperature or pressure because it is based on mass and not volume.

In this example, when 1 mole of the solute is dissolved in 1000 grams of naphthalene, the molality is calculated as follows:

• Convert the mass of naphthalene to kilograms: 1000 g = 1 kg.
• Use the formula: molality = moles of solute / mass of solvent in kg = 1 mole / 1 kg = 1 mol/kg.

This information is crucial for determining other properties such as the freezing point depression constant.

Freezing Point Depression Constant

The freezing point depression constant, denoted as Kf, is a property unique to each solvent. It indicates how much the freezing point of the solvent is lowered per molal concentration of a solute. In our exercise, we determine Kf for naphthalene using a solution with a known molality.

We use the formula \[\Delta T = Kf \times \text{molality} \times i\]where:

• \(\Delta T\) is the change in freezing point.
• molality is the solute concentration in mol/kg.
• \(i\) is the van't Hoff factor.

For naphthalene, given that it is a non-electrolyte, the van't Hoff factor (i) is 1. Here, \(\Delta T = 7°C\) with a molality of 1 mol/kg, so:\[Kf = \frac{7}{1 \times 1} = 7°C/\text{molal}\]This value helps us calculate freezing point depression for other solutes in naphthalene.

Van't Hoff Factor

The van't Hoff factor \(i\) is a term used in colligative properties to account for the effect of solute particle dissociation in solutions. It is the number of particles the solute splits into when dissolved. For non-electrolyte solutions, such as our naphthalene example, \(i\) is usually 1 because the solute does not dissociate in solution.

This factor is crucial when calculating colligative properties which involve freezing point depression. In this specific scenario with naphthalene and sulfur, sulfur does not dissociate, so \(i\) remains 1. Therefore, it does not alter the calculations of molality or the freezing point depression constant, simplifying the process of determining the molecular weight of the sulfur.