Problem 67
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
Verified Answer
The mole fraction of ethanol in the solution is \(\chi_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = \frac{\frac{1}{46.07}}{\frac{1}{18.015} + \frac{1}{46.07}}\). The vapor pressure of the solution at \(63.5^{\circ} \mathrm{C}\) is given by Raoult's Law: \(P_{\text{sol}} = \chi_{\mathrm{H}_{2}\mathrm{O}}P_{\mathrm{H}_{2}\mathrm{O}}^\circ + \chi_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}P_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}^\circ\). Finally, the mole fraction of ethanol in the vapor above the solution is found using Dalton's Law: \(\chi_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH, \text{vapor}}} = \frac{P_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}}{P_{\text{sol}}}\).
1Step 1: Calculate the moles of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)
Given equal masses of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), let's assume each has a mass of 1 gram. Calculate the number of moles of each:
For water: \(1 \; \text{g} \times \frac{1 \; \text{mol}}{18.015 \; \text{g/mol}}\)
For ethanol: \(1 \; \text{g} \times \frac{1 \; \text{mol}}{46.07 \; \text{g/mol}}\)
2Step 2: Calculate the mole fraction of ethanol in the solution
The mole fraction of ethanol can be found by dividing the moles of ethanol by the total number of moles in the solution:
\(\chi_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = \frac{\frac{1}{46.07}}{\frac{1}{18.015} + \frac{1}{46.07}}\)
3Step 3: Apply Raoult's Law to calculate the vapor pressure of the solution
Raoult's Law: \(P_{\text{sol}} = \chi_{\mathrm{H}_{2}\mathrm{O}}P_{\mathrm{H}_{2}\mathrm{O}}^\circ + \chi_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}P_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}^\circ\)
Where,
\(P_{\mathrm{H}_{2}\mathrm{O}}^\circ\) = vapor pressure of pure water = 175 torr
\(P_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}^\circ\) = vapor pressure of pure ethanol = 400 torr
\(\chi_{\mathrm{H}_{2}\mathrm{O}}\) = mole fraction of water in the solution
\(\chi_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}\) = mole fraction of ethanol in the solution (calculated in Step 2)
Plug in the values and calculate the vapor pressure of the solution.
4Step 4: Calculate the mole fraction of ethanol in the vapor
Using Dalton's Law, the ethanol vapor pressure can be found by multiplying the mole fraction of ethanol in the solution by the vapor pressure of the pure ethanol:
\(P_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = \chi_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}P_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}^\circ\)
Then, the mole fraction of ethanol in the vapor can be found by dividing the ethanol vapor pressure by the vapor pressure of the solution (from Step 3):
\(\chi_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH, \text{vapor}}} = \frac{P_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}}{P_{\text{sol}}}\)
Key Concepts
Vapor PressureMole FractionIdeal-Solution Behavior
Vapor Pressure
Vapor pressure is a measure of the tendency of molecules to escape from a liquid into the gaseous phase. It is a crucial concept when studying mixtures, particularly when predicting how volatile substances behave when mixed.
In the context of the exercise, vapor pressure is assessed at a specific temperature, which is 63.5°C. At this temperature, the pure vapor pressure of water (H₂O) is 175 torr, and that of ethanol (C₂H₅OH) is 400 torr.
When two substances are mixed, their individual vapor pressures contribute to the overall vapor pressure of the solution, assuming the solution behaves ideally. This contribution is determined using Raoult's Law, which helps calculate the total pressure exerted by the mixture."},{
In the context of the exercise, vapor pressure is assessed at a specific temperature, which is 63.5°C. At this temperature, the pure vapor pressure of water (H₂O) is 175 torr, and that of ethanol (C₂H₅OH) is 400 torr.
When two substances are mixed, their individual vapor pressures contribute to the overall vapor pressure of the solution, assuming the solution behaves ideally. This contribution is determined using Raoult's Law, which helps calculate the total pressure exerted by the mixture."},{
Mole Fraction
The mole fraction is a way of expressing the composition of a solution in terms of the proportion of moles of a particular component relative to the total number of moles in the solution. It is expressed as: \[ \chi_{component} = \frac{n_{component}}{n_{total}} \]Where \(n_{component}\) is the number of moles of the component of interest, and \(n_{total}\) is the total number of moles in the solution.
To find the mole fraction of ethanol in a solution where equal masses of water and ethanol are mixed, one needs to first calculate the moles of each substance from their masses. For our exercise, assuming each mass is 1 gram, the calculations are as follows:
\[ \chi_{C_{2}H_{5}OH} = \frac{\frac{1}{46.07}}{\frac{1}{18.015} + \frac{1}{46.07}} \]
This mole fraction is then used in further calculations, such as determining the vapor pressure of the solution using Raoult's Law.
To find the mole fraction of ethanol in a solution where equal masses of water and ethanol are mixed, one needs to first calculate the moles of each substance from their masses. For our exercise, assuming each mass is 1 gram, the calculations are as follows:
- Moles of water = \(\frac{1}{18.015}\)
- Moles of ethanol = \(\frac{1}{46.07}\)
\[ \chi_{C_{2}H_{5}OH} = \frac{\frac{1}{46.07}}{\frac{1}{18.015} + \frac{1}{46.07}} \]
This mole fraction is then used in further calculations, such as determining the vapor pressure of the solution using Raoult's Law.
Ideal-Solution Behavior
Ideal-solution behavior is an assumption where the solution's components mix perfectly without any interaction or change in the physical properties of the substances involved. It is a simplification that allows chemists to predict the behavior of the solution using Raoult's Law.
Raoult's Law mathematically defines the behavior of solutions under this ideal assumption. It states that the vapor pressure of an ideal solution is given by: \[ P_{sol} = \chi_{H_{2}O}P_{H_{2}O}^\circ + \chi_{C_{2}H_{5}OH}P_{C_{2}H_{5}OH}^\circ \]
This equation combines the mole fractions of each component with their corresponding pure vapor pressures to model the total vapor pressure of the solution.
In the exercise example, by multiplying the mole fractions by their respective vapor pressures (175 torr for water and 400 torr for ethanol), we obtain the total vapor pressure of the mixed solution at 63.5°C.
When the solution's behavior deviates from the ideal, calculations may require additional considerations about interactions between molecules, often difficult to quantify precisely in such simple models.
Raoult's Law mathematically defines the behavior of solutions under this ideal assumption. It states that the vapor pressure of an ideal solution is given by: \[ P_{sol} = \chi_{H_{2}O}P_{H_{2}O}^\circ + \chi_{C_{2}H_{5}OH}P_{C_{2}H_{5}OH}^\circ \]
This equation combines the mole fractions of each component with their corresponding pure vapor pressures to model the total vapor pressure of the solution.
In the exercise example, by multiplying the mole fractions by their respective vapor pressures (175 torr for water and 400 torr for ethanol), we obtain the total vapor pressure of the mixed solution at 63.5°C.
When the solution's behavior deviates from the ideal, calculations may require additional considerations about interactions between molecules, often difficult to quantify precisely in such simple models.
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