Problem 67 - Chemistry The Central Science

Question

At \(63.5^{\circ} \mathrm{C}\) the vapor pressure of \(\mathrm{H}_{2} \mathrm{O}\) is 175 torr, and that of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 400 torr. A solution is made by mixing equal masses of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} .\) (a) What is the mole fraction of ethanol in the solution? (b) Assuming ideal- solution behavior, what is the vapor pressure of the solution at \(63.5^{\circ} \mathrm{C} ?\) (c) What is the mole fraction of ethanol in the vapor above the solution?

Step-by-Step Solution

Answer

(a) The mole fraction of ethanol in the solution is 0.281. (b) The vapor pressure of the solution at 63.5°C is 236.7 torr. (c) The mole fraction of ethanol in the vapor above the solution is 0.472.

Step 1: Calculate the moles of H₂O and C₂H₅OH

First, we need to find the moles of H₂O and C₂H₅OH in the solution, since equal masses are mixed. We will use the molar mass of each substance to convert their masses to moles. molar mass of H₂O = 18.015 g/mol molar mass of C₂H₅OH = 46.07 g/mol Let's assume 46.07 grams of each substance is mixed, so the mass of the solution, m = 46.07 + 46.07 = 92.14 g. moles of H₂O = mass of H₂O / molar mass of H₂O = 46.07 g / 18.015 g/mol = 2.558 mol moles of C₂H₅OH = mass of C₂H₅OH / molar mass of C₂H₅OH = 46.07 g / 46.07 g/mol = 1 mol

Step 2: Calculate the mole fraction of ethanol in the solution

Now, let's calculate the mole fraction of ethanol (C₂H₅OH) in the solution using the moles calculated in Step 1. mole fraction of ethanol, x(C₂H₅OH) = moles of C₂H₅OH / (moles of H₂O + moles of C₂H₅OH) x(C₂H₅OH) = 1 mol / (2.558 mol + 1 mol) = 1/3.558 = 0.281 So, the mole fraction of ethanol in the solution is 0.281. Answer (a): The mole fraction of ethanol in the solution is 0.281.

Step 3: Calculate the vapor pressure of the solution using Raoult's Law

Now, we will calculate the vapor pressure of the solution at 63.5°C, assuming ideal-solution behavior. We can use Raoult's Law, which states that the vapor pressure of a substance in an ideal solution is proportional to its mole fraction. P(solution) = x(H₂O) * P₀(H₂O) + x(C₂H₅O) * P₀(C₂H₅OH) x(H₂O) = 1 - x(C₂H₅O) = 1 - 0.281 = 0.719 Now, we will plug in the given vapor pressures of water and ethanol. P₀(H₂O) = 175 torr P₀(C₂H₅OH) = 400 torr P(solution) = (0.719 * 175 torr) + (0.281 * 400 torr) ≈ 124.3 torr + 112.4 torr = 236.7 torr Answer (b): The vapor pressure of the solution at 63.5°C is 236.7 torr.

Step 4: Calculate the mole fraction of ethanol in the vapor above the solution

To find the mole fraction of ethanol in the vapor above the solution, we will use the partial pressures of H₂O and C₂H₅OH in the vapor. P(H₂O) = x(H₂O) * P₀(H₂O) = 0.719 * 175 torr ≈ 125.8 torr P(C₂H₅OH) = x(C₂H₅OH) * P₀(C₂H₅OH) = 0.281 * 400 torr ≈ 112.4 torr x(vapor, C₂H₅OH) = P(C₂H₅OH) / (P(H₂O) + P(C₂H₅OH)) x(vapor, C₂H₅OH) = 112.4 torr / (125.8 torr + 112.4 torr) ≈ 0.472 Answer (c): The mole fraction of ethanol in the vapor above the solution is 0.472.

Key Concepts

Raoult's Law

Raoult's Law is a fundamental principle in physical chemistry that describes how the vapor pressure of a solution component relates to its concentration. According to Raoult's Law, the vapor pressure of a volatile component in an ideal solution is directly proportional to its mole fraction. In mathematical terms, the law is expressed as:
\[ P_i = x_i \times P_i^0 \]
where \( P_i \) is the partial vapor pressure of component \( i \) in the solution, \( x_i \) is the mole fraction of component \( i \), and \( P_i^0 \) is the pure component's vapor pressure at the same temperature.
In the given exercise, we applied Raoult's Law to calculate the overall vapor pressure of a solution made by mixing water and ethanol in equal masses. By knowing the mole fraction of each component and their pure vapor pressures, we were able to determine the solution's vapor pressure at a specific temperature. This illustrates the practical application of Raoult's Law in deriving properties of solutions from its individual components.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the number of moles of a particular substance to the total number of moles of all substances present in the mixture. The formula to calculate the mole fraction (\( x_i \)) of a substance \( i \) is given by:
\[ x_i = \frac{n_i}{\text{total number of moles in solution}} = \frac{n_i}{\text{\( n_1 + n_2 + \text{...} + n_i + \text{...} + n_n \)}} \]
where \( n_i \) is the number of moles of substance \( i \) and \( n_1, n_2, ..., n_n \) are the moles of other substances in the solution. In the exercise, the students must understand how to convert the equal masses of water and ethanol to moles before applying this formula. By determining the mole fraction of ethanol, we have a crucial piece of information to utilize Raoult's Law for calculating the vapor pressures.

Ideal Solution Behavior

Ideal solution behavior is an important concept when dealing with solutions. It refers to a solution where the intermolecular forces between like and unlike molecules are equal, resulting in behavior that adheres closely to Raoult's Law throughout the entire range of concentrations. Ideal solutions exhibit a linear correlation between the mole fraction and vapor pressure of their components. In such solutions, there is no volume change upon mixing and the enthalpy of mixing is zero, meaning no heat is absorbed or evolved.This concept is crucial in our exercise, where we assume that the water-ethanol mixture behaves as an ideal solution, simplifying our calculations of vapor pressures and mole fractions. However, students should keep in mind that in reality, most solutions deviate from ideal behavior, and these deviations need to be taken into account for accurate predictions in practical applications.