Question

A solution contains 3.75 g of a nonvolatile pure hydrocarbon in 95 g acetone. The boiling points of pure acetone and the solution are \(55.95^{\circ} \mathrm{C}\) and \(56.50^{\circ} \mathrm{C},\) respectively. The molal boiling- point constant of acetone is \(1.71^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\) . What is the molar mass of the hydrocarbon?

Short answer

The molar mass of the hydrocarbon is approximately 122.92 g/mol. To find this, we used the boiling-point elevation formula, calculated the change in boiling point (ΔT), determined the molality of the solution, calculated the moles of solute (hydrocarbon), and finally divided the mass of the hydrocarbon by the moles of the hydrocarbon.

Step-by-step solution

Understand the boiling-point elevation formula

The formula for boiling-point elevation is: ΔT = Kb * molality Where ΔT is the change in boiling point due to the presence of solute, Kb is the molal boiling-point constant of the solvent, and molality is the molality of the solution (mol solute per kg of solvent). We are given the boiling points of the pure acetone and the solution, and the molal boiling-point constant (Kb) of acetone. With this information, we will be able to calculate the molality of the solution.

Calculate the change in boiling point (ΔT)

To calculate ΔT, we subtract the boiling point of pure acetone from the boiling point of the solution: ΔT = Boiling point of the solution - Boiling point of pure acetone ΔT = 56.50 °C - 55.95 °C = 0.55 °C

Calculate the molality of the solution

We use the boiling-point elevation formula to calculate the molality of the solution: Molality = ΔT / Kb Molality = 0.55 °C / 1.71 °C⋅kg/mol = 0.3216 mol/kg

Calculate the moles of solute (hydrocarbon)

As we have the molality of the solution and the amount of solvent in kg, we can calculate the moles of solute (hydrocarbon) present in the solution: Moles of solute = Molality * (mass of solvent in kg) Moles of solute = 0.3216 mol/kg * (95 g / 1000 g/kg) = 0.03052 mol

Calculate the molar mass of the hydrocarbon

Finally, we can calculate the molar mass of the hydrocarbon by dividing the mass of the hydrocarbon by the moles of the hydrocarbon: Molar mass = mass of hydrocarbon / moles of hydrocarbon Molar mass = 3.75 g / 0.03052 mol ≈ 122.92 g/mol The molar mass of the hydrocarbon is approximately 122.92 g/mol.

Key concepts

Molal Boiling-Point Constant

The molal boiling-point constant, often represented as \(K_b\), plays a crucial role in understanding how solutes affect boiling points. When a nonvolatile solute is added to a solvent, it causes an elevation in the boiling point of the solution. This occurrence is known as boiling-point elevation. The constant \(K_b\) quantifies how much the boiling point of a solvent is elevated per molal concentration of the solute.
The concept can be summarized by the formula:

• \( \Delta T = K_b \times \text{molality} \)

\( \Delta T \) is the change in temperature, \(K_b\) is the molal boiling-point constant, and molality is the amount of solute per kg of solvent. In this particular problem, the molal boiling-point constant of acetone is given as \(1.71^{\circ} \text{C} \cdot \text{kg/mol}\), showing how sensitive acetone is to changes caused by solutes.

Molar Mass Calculation

Calculating the molar mass of a substance involves determining how much one mole of that substance weighs in grams. This task often involves using experimental data like change in boiling point when a solute is added to a solvent. In the provided exercise, the molar mass was found using the relationship between mass, molality, and the change in boiling point.
Molar mass can be calculated using the formula:

• \( \text{Molar Mass} = \frac{\text{mass of solute}}{\text{moles of solute}} \)

Here, the mass of the hydrocarbon is 3.75 grams, and through calculations involving the acetone's \(K_b\) and boiling-point change, the moles of solute were found to be approximately 0.03052 mol. Thus, the molar mass of the hydrocarbon is calculated to be roughly 122.92 g/mol. This process showcases how physical properties can be leveraged to discern the characteristics of solutes in solution chemistry.

Solution Chemistry

Solution chemistry focuses on understanding how substances interact and dissolve in solvents to form solutions. This field studies properties such as concentration, solubility, and changes in boiling and freezing points. Solutions consist of a solvent (in larger quantity) and one or more solutes (in smaller quantities).
In the context of this exercise, a nonvolatile hydrocarbon solute is dissolved in acetone, the solvent. As the solute dissolves, it disrupts the solvent's normal boiling behavior, causing an elevation in the boiling point. This principle helps identify properties like molar masses of unknown solutes.
Boiling-point elevation is just one of many colligative properties studied in solution chemistry, which are unique as they depend on the number of solute particles rather than the nature of the particles themselves. Understanding these concepts provides valuable insights into the behavior of mixtures and the effects of solutes on solvents.