Problem 102 - Chemistry: The Central Science in SI Units, Global Edition

Question

The normal boiling point of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\) is \(78.4^{\circ} \mathrm{C}\). When \(3.26 \mathrm{~g}\) of a soluble nonelectrolyte is dissolved in \(100.0 \mathrm{~g}\) of ethanol at that temperature, the vapor pressure of the solution is \(100 \mathrm{kPa}\). What is the molar mass of the solute?

Step-by-Step Solution

Answer

The molar mass of the solute cannot be determined with the given information because, when using Raoult's Law and the molality of the solution, we end up with an indeterminate form due to the mole fraction of the solute being 0. There might be an error in the given information or a small deviation in the mole fraction.

Step 1: Write down Raoult's Law formula and define variables

According to Raoult's law, the vapor pressure of a solution equals the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. \[P = X_A P^*_A\] Where: \(P\) = vapor pressure of the solution \(P^*_A\) = vapor pressure of the pure solvent \(X_A\) = mole fraction of the solvent

Step 2: Calculate mole fraction of the solvent

First, we convert the vapor pressure of the solution to the same units as the boiling point of ethanol. \[\text{Vapor Pressure of the solution} = 100\, kPa = 760 \, mmHg \] Next, we will calculate the mole fraction of the solvent. Rearrange Raoult's law formula to find \(X_A\): \[X_A = \frac{P}{P^*_A}\] Now, we plug in the given values: \[X_A = \frac{760\,mmHg}{760\,mmHg} = 1\] So the mole fraction of the solvent is 1, which means that the mole fraction of the solute is 0.

Step 3: Calculate molality of the solution

Now we can calculate the molality (m) of the solution using the mass of solute dissolved in the solvent. \[m = \frac{moles \,of \,solute}{kg\,of\, solvent}\] We are given: - mass of solute: 3.26 g - mass of solvent (ethanol): 100.0 g = 0.100 kg To find the moles of solute, we need to find the molar mass of the solute. Let's represent the molar mass of solute by "M". \[moles \,of\,solute = \frac{3.26\,g}{M}\] Plug in the given values: \[m = \frac{\frac{3.26\,g}{M}}{0.100\,kg}\]

Step 4: Use Raoult's Law and Molality to find Molar Mass

We know that the mole fraction of the solute is 0, so the molality of the solution can also be represented as: \[m = \frac{moles \,of \,solute}{moles \,of \,solvent}\] Now, we can use the mole fraction and molality of the solution to find the molar mass of solute (M): \(m = \frac{\frac{3.26\,g}{M}}{0.100\,kg} = \frac{moles \,of \,solute}{moles \,of \,solvent}\) Solving for M: \[M = \frac{3.26\,g \times moles \,of \,solvent}{0.100\,kg \times moles \,of \,solute}\] Now let's plug in the mole fraction of solute, which is 0: \[M = \frac{3.26\,g \times 0}{0.100\,kg \times 0}\] Since the mole fraction of solute is 0, we get an indeterminate form. However, we know that the mole fraction of solute cannot be exactly zero because there is a presence of solute in the solution. Thus, there might be an error in the given information or a small deviation in the mole fraction. Therefore, we cannot determine the molar mass of the solute with the given information.

Key Concepts

Boiling Point

The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. For ethanol, this happens at 78.4°C under standard atmospheric pressure, which is about 760 mmHg. This means that at this temperature, the ethanol molecules have enough energy to escape the liquid and become gas.

The boiling point can change if the liquid is not pure. When we dissolve a substance (like a nonelectrolyte solute) into a solvent, the boiling point usually increases due to the presence of solute particles. This is known as ebullioscopic boiling point elevation. This principle is important when working with solutions as it helps us to understand how solutes affect the boiling characteristics of a solvent.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase. When a solute is added to a solvent, it typically lowers the solvent's vapor pressure, as the solute particles take up space at the surface and inhibit some solvent molecules from escaping into the vapor phase.

In this exercise, Raoult's Law is key. It relates the vapor pressure of a solution to the mole fractions of its components. The formula states:

• \( P = X_A P^*_A \)

Here, \(P\) is the vapor pressure of the solution, \(P^*_A\) is the vapor pressure of the pure solvent, and \(X_A\) is the mole fraction of the solvent. In simpler terms, for a solution to have lower vapor pressure than the pure solvent, the presence of a solute disperses some of the surface molecules, preventing them from evaporating as easily.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a solution. It is defined as the ratio of the number of moles of a component to the total number of moles in the solution. This is a dimensionless number that helps in understanding the relative amounts of different substances in a mixture.

To find the mole fraction of the solvent in a solution, using Raoult's Law, you need to know the vapor pressure of both the solution and the pure solvent. Using the formula derived from Raoult's Law:

• \( X_A = \frac{P}{P^*_A} \)

When you substitute the values correctly, you will be able to determine the relative mole fractions of the substance in a mixture, which can give insights into the solution's properties and behaviors.

Molar Mass Calculation

Calculating the molar mass of a solute in a solution involves understanding the relationship between the mass of the solute, the number of moles it represents, and its effect on the solution's properties like vapor pressure and boiling point.

The molar mass \(M\) is calculated using the formula:

• \( m = \frac{moles \; of \; solute}{mass \; of \; solvent} \)

When rearranged to solve for \(M\), it allows the identification of the solute's molar mass given sufficient experimental data. Understanding this calculation is vital for characterizing unknown substances dissolved in a solution. However, in the context of this exercise, if a miscalculation or incorrect assumptions about the mole fraction occur, accurately finding the solute's molar mass becomes challenging. This highlights the importance of precise and correct information in chemical calculations.