Problem 68
Question
At \(20^{\circ} \mathrm{C}\) the vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 75 torr, and that of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is 22 torr. Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fractions of a solution that has a vapor pressure of 35 torr at \(20^{\circ} \mathrm{C} ?\) (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?
Step-by-Step Solution
Verified Answer
The composition of the solution in mole fractions is \(X_1 = \frac{13}{53}\) for benzene and \(X_2 = \frac{40}{53}\) for toluene. The mole fraction of benzene in the vapor above the solution is 0.528.
1Step 1: Understanding Raoult's Law
Raoult's Law states that for an ideal solution, the vapor pressure of the component (i) in the solution is directly proportional to its mole fraction (Xi) in the solution:
\[p_i = X_i \times p_i^*\]
Where:
\(p_i\) = vapor pressure of the component (i) in the solution
\(X_i\) = mole fraction of component (i) in the solution
\(p_i^*\) = vapor pressure of the pure component (i) at the same temperature
In this problem, we have benzene (C6H6) and toluene (C7H8).
2Step 2: Write the equations for benzene and toluene using Raoult's law
Let the mole fraction of benzene (C6H6) in the solution be X1 and that of toluene (C7H8) be X2.
For benzene:
\[p_1 = X_1 \times p_1^*\]
For toluene:
\[p_2 = X_2 \times p_2^*\]
Where:
\(p_1\) = vapor pressure of benzene in the solution
\(p_2\) = vapor pressure of toluene in the solution
\(p_1^*\) = vapor pressure of pure benzene at 20°C = 75 torr
\(p_2^*\) = vapor pressure of pure toluene at 20°C = 22 torr
3Step 3: Find the total vapor pressure and set up the mole fraction equation
The total vapor pressure of the solution is given as 35 torr. It is equal to the sum of the individual vapor pressures of benzene and toluene.
\[p_{total} = p_1 + p_2\]
Since it is an ideal solution, the mole fractions of benzene and toluene must add up to 1:
\[X_1 + X_2 = 1\]
4Step 4: Solve the equations for mole fractions of benzene and toluene
Let's substitute the equations from step 2 into the total vapor pressure equation:
\[35 = X_1 \times 75 + X_2 \times 22\]
Now substitute the mole fraction equation from step 3 into the equation above:
\[35 = X_1 \times 75 + (1-X_1) \times 22\]
Now, we can solve for the mole fraction of benzene (X1):
\(35 = 75X_1 + 22 - 22X_1\)
\(35 - 22 = 75X_1 - 22X_1\)
\(13 = 53X_1\)
\(X_1 = \frac{13}{53}\)
Now, we can find the mole fraction of toluene (X2) using the mole fraction equation from step 3:
\(X_2 = 1 - X_1\)
\(X_2 = 1 - \frac{13}{53}\)
\(X_2 = \frac{40}{53}\)
The composition of the solution in mole fractions is \(X_1 = \frac{13}{53}\) for benzene and \(X_2 = \frac{40}{53}\) for toluene.
5Step 5: Find the mole fraction of benzene in the vapor above the solution
Now that we have the mole fraction of benzene and toluene in the solution, we can use Raoult's law to find the mole fraction of benzene in the vapor.
Recall the equation for benzene from Step 2:
\[p_1 = X_1 \times p_1^*\]
Substitute the values:
\(p_1 = \left(\frac{13}{53}\right) \times 75\)
\(p_1 = 18.49\,\text{torr}\)
Now, we can find the mole fraction of benzene in the vapor (Y1) using the following relation:
\[Y_1 = \frac{p_1}{p_{total}}\]
Substitute the values:
\(Y_1 = \frac{18.49}{35}\)
\(Y_1 = 0.528\)
The mole fraction of benzene in the vapor above the solution is 0.528.
Key Concepts
Vapor PressureIdeal SolutionMole Fraction
Vapor Pressure
Vapor pressure is an important concept in understanding how liquids behave under different conditions. It refers to the pressure exerted by the vapor when a liquid is in a closed system at a given temperature. In simpler terms, it's like the pushing force produced by evaporated molecules when they escape from the liquid's surface into the air above. This force depends on the nature of the liquid and the surrounding temperature.
Understanding vapor pressure is crucial for predicting how a solution behaves, especially when determining the boiling point or understanding evaporation rates.
- Higher temperatures increase kinetic energy, making molecules escape easier, and thus raising the vapor pressure.
- Different substances have different intrinsic vapor pressures due to variations in molecular structure and bond strength.
Understanding vapor pressure is crucial for predicting how a solution behaves, especially when determining the boiling point or understanding evaporation rates.
Ideal Solution
An ideal solution is a theoretical concept where the interactions between different molecules of the components in a mixture are similar to the interactions between the molecules of each component in its pure state. This means that in an ideal solution, the forces between like and unlike molecules are the same. Consequently, when mixed in any proportion, these substances behave predictably, adhering closely to Raoult's Law.
- This assumption simplifies calculations, making it easier to predict properties like vapor pressure.
- Carbon-based solutions like benzene and toluene often approach ideality, particularly when the molecular sizes and intermolecular forces are similar.
Mole Fraction
The mole fraction is a way of expressing the concentration of a component in a mixture. It is calculated by dividing the number of moles of one component by the total number of moles of all components in the mixture. It's a dimensionless number between 0 and 1, showing the proportion of one component relative to the entire mix.
- A mole fraction close to 1 implies the component dominates the mixture.
- Conversely, values closer to 0 indicate that the component is a minor part of the solution.
Other exercises in this chapter
Problem 65
(a) Calculate the vapor pressure of water above a solution prepared by adding \(22.5 \mathrm{~g}\) of lactose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}
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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} \mathr
View solution Problem 69
(a) Why does a \(0.10 m\) aqueous solution of NaCl have a higher boiling point than a \(0.10 \mathrm{~m}\) aqueous solution of \(\mathrm{C}_{6} \mathrm{H}_{12}
View solution Problem 70
Arrange the following aqueous solutions, each \(10 \%\) by mass in solute, in order of increasing boiling point: glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12}
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