Problem 102 - Chemistry The Central Science

Question

The normal boiling point of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\) is \(78.4^{\circ} \mathrm{C} .\) When 9.15 \(\mathrm{g}\) of a soluble nonelectrolyte is dissolved in 100.0 \(\mathrm{g}\) of ethanol at that temperature, the vapor pressure of the solution is \(7.40 \times 10^{2}\) torr. What is the molar mass of the solute?

Step-by-Step Solution

Answer

The molar mass of the solute is 348 g/mol.

Step 1: Determine the molality of the solution

(using Raoult's Law) Raoult's Law states that the vapor pressure of a solution (P) is equal to the vapor pressure of the pure solvent (P°) times the mole fraction of the solvent (X₁): \[ P = P° * X₁ \]

Step 2: Calculate the mole fraction of the solvent

Given: Vapor pressure of the solution (P) = \(7.40 \times 10^2\) torr Normal boiling point (Vapor pressure) of ethanol (P°) = 78.4°C (The normal boiling point is the temperature at which the vapor pressure of the substance in its liquid phase is equal to 1 atm or 760 torr) First, convert the normal boiling point to torr: P° = 78.4°C = (78.4°C + 273.15) K = 351.55 K Vapor pressure of pure ethanol = 760 torr (at its boiling point) So now we can find the mole fraction of the solvent (X₁ = P/P°): X₁ = P/P° = (\(7.40 \times 10^2\) torr) / (760 torr) = 0.9737

Step 3: Calculate the mole fraction of the solute

Mole fraction of the solute (X₂) can be calculated as follows: X₂ = 1 - X₁ = 1 - 0.9737 = 0.0263

Step 4: Calculate the molality of the solution

Molality (m) can be calculated as: m = X₂ * (1000 g/kg) / (mass of solvent in kg) Since we have 100 g of ethanol as solvent, we can convert it to kg: mass of solvent (ethanol) = 100.0 g = 0.1 kg Now calculate the molality: m = 0.0263 * (1000 / 0.1) = 0.0263 * 10000 = 263 mol/kg

Step 5: Calculate the molar mass of the solute

Given that 9.15 g of solute is dissolved in 0.1 kg of ethanol, we can use the formula: Molar mass of solute = mass of solute / (molality * mass of solvent in kg) Molar mass of solute = 9.15 g / (263 mol/kg * 0.1 kg) = 9.15 g / 26.3 mol = 348 g/mol Hence, the molar mass of the solute is 348 g/mol.

Key Concepts

Raoult's Law

Raoult's Law is a principle used to determine the vapor pressure of solutions. It states that the vapor pressure of a solution is the product of the mole fraction of the solvent and the vapor pressure of the pure solvent. In a formula, this law is expressed as \( P = P^\circ \times X_1 \). Here, \( P \) represents the vapor pressure of the solution, \( P^\circ \) is the vapor pressure of the pure solvent, and \( X_1 \) is the mole fraction of the solvent. This concept helps in understanding how a solute affects the boiling point and vapor pressure when dissolved in a solvent.
To utilize Raoult's Law, it's important to accurately find the mole fraction of the solvent. This involves comparing the vapor pressures (both measured and of the pure state). Note however, this applies to ideal solutions where interactions between molecules are similar to the solvent alone.

Boiling Point

The boiling point of a substance is the temperature at which its vapor pressure equals the external pressure. For liquids like ethanol, knowing the boiling point is crucial in calculating changes when solutes dissolve in it. At the boiling point, liquids turn to vapor. When a solute is added, it causes the boiling point to elevate, requiring a higher temperature for boiling.
In our example, ethanol's normal boiling point at one atmosphere pressure is 78.4°C, corresponding to a vapor pressure of 760 torr. Using Raoult's Law, the impact of adding non-volatile solute, such as the one in our problem, can be seen as it affects the solution's properties, increasing the boiling point.

Molar Mass Calculation

Calculating molar mass is an essential part of chemistry. It helps determine the amount of substance needed or produced in a reaction. Molar mass is the mass of one mole of a given substance and is expressed in grams per mole (g/mol).
Following the steps, you use the molality derived from the mole fraction of the solute and the known mass of the solute. In the problem, the molar mass is found using the formula: Molar mass of solute = \( \frac{\text{mass of solute}}{\text{molality} \times \text{mass of solvent in kg}} \).

• The molality is found through the mole fractions calculated using Raoult's Law.
• Knowing the mass of the solute, you compute the molar mass.
• In our context, the molar mass calculation confirms the solute's properties and identity as 348 g/mol.

This highlights the interplay between vapor pressure, boiling point, and molecular composition.