Question

A \(1.60-\mathrm{g}\) sample of a mixture of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) and anthracene \(\left(\mathrm{C}_{14} \mathrm{H}_{10}\right)\) is dissolved in \(20.0 \mathrm{~g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\). The freezing point of the solution is \(2.81^{\circ} \mathrm{C}\). What is the composition as mass percent of the sample mixture? The freezing point of benzene is \(5.51^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(5.12^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\).

Short answer

The composition of the sample mixture is approximately 15.625 % naphthalene and 84.375 % anthracene.

Step-by-step solution

Calculate the Freezing Point Depression

Given the freezing point of the solution and the freezing point of pure benzene, we can find the freezing point depression (∆Tf) using the following formula: ∆Tf = Tf, pure - Tf, solution ∆Tf = \(5.51^{\circ}\mathrm{C} - 2.81^{\circ}\mathrm{C} = 2.70^{\circ}\mathrm{C}\)

Use Freezing Point Depression and Kf to Calculate Molality

We can use the freezing point depression constant (Kf) to find the molality of the solute. The formula is: ∆Tf = Kf * molality Rearrange the formula to solve for molality: molality = ∆Tf / Kf molality = \(2.70^{\circ}\mathrm{C} / 5.12^{\circ}\mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) molality = 0.5273 mol/kg

Calculate Moles of Solute

Using the molality and the mass of benzene, we can calculate the moles of solute: moles of solute = molality * mass of solvent (in kg) moles of solute = 0.5273 mol/kg * 0.020 kg moles of solute = 0.010546 mol

Set Up System of Equations

Let x be the mass of naphthalene and y be the mass of anthracene in the mixture. Then we have: x + y = 1.60 g (Total mass of mixture) We also know that the total moles of solute is 0.010546 mol. Therefore: (x / molecular weight of naphthalene) + (y / molecular weight of anthracene) = 0.010546

Solve System of Equations

Using the molecular weights (128.17 g/mol for naphthalene and 178.23 g/mol for anthracene), we can solve the system of equations: x / 128.17 + y / 178.23 = 0.010546 We already know x + y = 1.60, so we can use substitution or elimination to solve for x and y: x = 1.60 - y Substitute: (1.60 - y) / 128.17 + y / 178.23 = 0.010546 Now you can solve for y: y ≈ 1.35 g (mass of anthracene) And x: x ≈ 0.25 g (mass of naphthalene)

Calculate Mass Percent

Finally, we can calculate the mass percent of each component: Mass percent of naphthalene = (mass of naphthalene / total mass) * 100 = (0.25 g / 1.60 g) * 100 = 15.625 % Mass percent of anthracene = (mass of anthracene / total mass) * 100 = (1.35 g / 1.60 g) * 100 = 84.375 % Thus, the composition of the sample mixture is approximately 15.625 % naphthalene and 84.375 % anthracene.

Key concepts

Molality Calculation

Understanding molality calculation is critical in the study of solutions and their properties. Molality (m) is defined as the number of moles of solute per kilogram of solvent. It is expressed using the unit moles per kilogram (\texttt{mol/kg}). This measure is preferable in some cases because unlike molarity, molality does not change with temperature since mass remains constant.

To calculate the molality, one can use the following formula:
\[\text{molality} = \frac{\text{number of moles of solute}}{\text{mass of solvent (in kg)}}\]
This exercise demonstrates a practical use of molality calculation. Here, the molality is found by measuring the freezing point depression (the drop in freezing point compared to the pure solvent) and dividing it by the cryoscopic constant (Kf), which is a property specific to the solvent. This step is crucial for determining the amount of solute in terms of its molar quantity and is key in carrying out subsequent steps like molecular weight determination.

Colligative Properties

The colligative properties of a solution depend on the concentration of solute particles, regardless of their identity. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure.

Freezing point depression is one colligative property observed when a solute is dissolved in a solvent, resulting in the lowering of its freezing point. It can be calculated using the formula:
\[\Delta Tf = Kf \times \text{molality}\]
Here, \(\Delta Tf\) is the change in freezing point, and Kf is the cryoscopic constant of the solvent.

For students working on exercises related to freezing point depression, understanding this colligative property helps in making accurate calculations for solute concentrations. Especially in this scenario, knowing the freezing point depression allows for indirect analysis of the mixture's components by linking changes in the physical property to the amount of solute.

Molecular Weight Determination

Determining the molecular weight is essential in analyzing the composition of chemical substances. In the provided exercise, it aids in finding the mass percentages of two solutes in a binary mixture. The molecular weights of the individual components are used to establish relationships between mass and moles, and they serve as conversion factors.

From the molality and the freezing point depression data, we calculate the moles of the mixture. Then, by setting up a system of equations relating the masses of the individual components (naphthalene and anthracene) to their molecular weights, we can deduce the exact composition of the mixture.

This method is a classic example of how molecular weight determination can be practical and powerful. Without directly measuring the individual quantities, molecular weight permits a back-calculation from observable physical changes, such as freezing point depression, to the proportions of each compound in the mixture.