Question

At \(63.5^{\circ} \mathrm{C}\), the vapor pressure of \(\mathrm{H}_{2} \mathrm{O}\) is \(23.3 \mathrm{kPa}\), and that of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is \(53.3 \mathrm{kPa}\). 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 idealsolution behavior, what is the vapor pressure of the solution at \(63.5^{\circ} \mathrm{C} ?(\mathbf{c})\) What is the mole fraction of ethanol in the vapor above the solution?

Short answer

(a) The mole fraction of ethanol in the solution is \(0.2815\). (b) The vapor pressure of the solution at \(63.5^{\circ}C\) is \(31.74\,\text{kPa}\). (c) The mole fraction of ethanol in the vapor above the solution is \(0.473\).

Step-by-step solution

- Calculate the moles of water and ethanol

First, we are given equal masses of water and ethanol. We can find the moles of each substance by dividing the mass by the molar mass: For water (\(H_2O\)): Molar mass of \(H_2O = 18.02\,g/mol\) For ethanol (\(C_2H_5OH\)): Molar mass of \(C_2H_5OH = 46.07\,g/mol\) Let's assume we have 1 gram of each substance. Then, the moles of water and ethanol can be calculated as follows: Moles of \(H_2O = \frac{1}{18.02} = 0.0555\,mol\) Moles of \(C_2H_5OH =\frac{1}{46.07} = 0.0217\,mol\)

- Calculate the mole fractions of water and ethanol in the solution

Now, we will calculate the mole fractions (\(x_i\)) of water and ethanol in the solution. The mole fraction of each component can be calculated using the formula: \(x_i = \frac{\text{moles of component i}}{\text{total moles in the solution}}\) Total moles in the solution = moles of \(H_2O\) + moles of \(C_2H_5OH = 0.0555 + 0.0217 = 0.0772\,mol\) Mole fraction of \(H_2O \,(x_{H_2O}) = \frac{0.0555\,\text{mol}}{0.0772\,\text{mol}} = 0.7185\) Mole fraction of \(C_2H_5OH \,(x_{C_2H_5OH}) = \frac{0.0217\,\text{mol}}{0.0772\,\text{mol}} = 0.2815\) (a) The mole fraction of ethanol in the solution is \(0.2815\).

- Calculate the vapor pressure of the solution using Raoult's Law

As we are assuming ideal solution behavior, we can use Raoult's Law to calculate the vapor pressure of the solution. Raoult's Law states that the partial vapor pressure of a component in an ideal solution is equal to the product of the mole fraction of the component in the solution and its vapor pressure in the pure state: Partial vapor pressure of \(H_2O \,(P_{H_2O}) = x_{H_2O} \times P^{0}_{H_2O} = 0.7185 \times 23.3\,\text{kPa} = 16.73\,\text{kPa}\) Partial vapor pressure of \(C_2H_5OH \,(P_{C_2H_5OH}) = x_{C_2H_5OH} \times P^{0}_{C_2H_5OH} = 0.2815 \times 53.3\,\text{kPa} = 15.01\,\text{kPa}\) The total vapor pressure of the solution can be obtained by adding the partial vapor pressures: Total vapor pressure of the solution \(P = P_{H_2O} + P_{C_2H_5OH} = 16.73 + 15.01 = 31.74\,\text{kPa}\) (b) The vapor pressure of the solution at \(63.5^{\circ}C\) is \(31.74\,\text{kPa}\).

- 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 (\(y_{C_2H_5OH}\)), we can use the formula: \(y_{C_2H_5OH} = \frac{P_{C_2H_5OH}}{P} = \frac{15.01\,\text{kPa}}{31.74\,\text{kPa}} = 0.473\) (c) The mole fraction of ethanol in the vapor above the solution is \(0.473\).

Key concepts

Vapor Pressure

Vapor pressure is a crucial concept in the study of solutions and lies at the heart of Raoult's Law. It is defined as the pressure exerted by the vapor when a liquid is in a closed system. This means it represents the force the vapor applies against the atmosphere when it is in balance.
Given that the vapor pressure depends on temperature, it is influenced by how easily a liquid's molecules can escape into the vapor phase.
Understanding vapor pressure is key because it tells us about the volatility of a substance. Substances with high vapor pressures are more volatile, meaning they evaporate more easily.

• Pure substances like water and ethanol have their own characteristic vapor pressures at a given temperature.
• In a solution, each component's vapor pressure gets influenced by its presence within the mixture.

Knowing the vapor pressure of a solution helps predict how it behaves when exposed to different temperatures and conditions.

Mole Fraction

The concept of the mole fraction is essential for understanding mixtures in chemistry, specifically when dealing with solutions. The mole fraction provides us with the proportion of moles of a component to the total moles in the solution.
This ratio gives insight into how much of each component is present.
To calculate the mole fraction (\(x_i\)) of a component, use:

\(x_i = \frac{\text{moles of component i}}{\text{total moles in the solution}}\)

With a mole fraction, it becomes easier to determine the concentration of a substance without the need for a scale. The sum of the mole fractions for all components in a solution equals 1.
For example, in a solution of water and ethanol, knowing the mole fraction of each gives insight into the prevalence of one component over another.

• This serves as a building block for further calculations, such as determining the partial vapor pressures of each component in a solution.
• Mole fractions remain constant as long as the molar ratios in the solution don't change, making it a stable measure of concentration.

Ideal Solution

An ideal solution is a key concept in liquid mixtures, referring to a situation where the intermolecular forces between molecules of different components are the same as those between molecules of the same component. This means the interactions in the mixture are uniform, and there is no change in energy upon mixing.
The hallmark of ideal solutions is that their behavior can be accurately described by Raoult's Law. This law states that the partial vapor pressure of each component in an ideal solution is directly proportional to its mole fraction.

• Ideal solutions assume that properties like volume and enthalpy remain unchanged upon mixing.
• While many real-world solutions deviate slightly, some systems, like benzene and toluene, approximate ideal behavior quite closely.
• Understanding ideal solutions helps us predict how a mixture will behave without complex calculations or experimental determination.

In our scenario, the assumption of ideal solution behavior allowed us to use Raoult's Law effectively to calculate the vapor pressures and the related mole fractions in the vapor phase.