To solve the problem of calculating the total pressures \( P_A \) and \( P_B \) for the two ideal solutions formed by benzene and toluene, we first need to understand how to apply Raoult's Law. This law states that the partial vapor pressure of each component in an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
Step-by-Step Calculation
1. Analyzing Solution A
Solution A contains equal moles of benzene and toluene. Let's denote the number of moles of each component as \( n \). Therefore, the total number of moles in solution A is:
- Total moles = \( n_{\text{benzene}} + n_{\text{toluene}} = n + n = 2n \)
The mole fractions can be calculated as follows:
- Mole fraction of benzene, \( X_{\text{benzene}} = \frac{n}{2n} = \frac{1}{2} \)
- Mole fraction of toluene, \( X_{\text{toluene}} = \frac{n}{2n} = \frac{1}{2} \)
Using Raoult's Law, the partial pressures are:
- Partial pressure of benzene, \( P_{\text{benzene}} = X_{\text{benzene}} \times P^{\circ}_{\text{benzene}} = \frac{1}{2} \times 160 \, \text{mm Hg} = 80 \, \text{mm Hg} \)
- Partial pressure of toluene, \( P_{\text{toluene}} = X_{\text{toluene}} \times P^{\circ}_{\text{toluene}} = \frac{1}{2} \times 60 \, \text{mm Hg} = 30 \, \text{mm Hg} \)
The total pressure \( P_A \) for solution A is the sum of the partial pressures:
- \( P_A = P_{\text{benzene}} + P_{\text{toluene}} = 80 + 30 = 110 \, \text{mm Hg} \)
2. Analyzing Solution B
Solution B contains equal masses of benzene and toluene. Let's assume we have \( m \) grams of each component. The molar masses are approximately:
- Molar mass of benzene (C₆H₆) = 78 g/mol
- Molar mass of toluene (C₇H₈) = 92 g/mol
The number of moles for each component can be calculated as:
- Moles of benzene, \( n_{\text{benzene}} = \frac{m}{78} \)
- Moles of toluene, \( n_{\text{toluene}} = \frac{m}{92} \)
The total number of moles in solution B is:
- Total moles = \( n_{\text{benzene}} + n_{\text{toluene}} = \frac{m}{78} + \frac{m}{92} \)
To find a common denominator, we can calculate:
- Common denominator = 78 × 92 = 7176
- Thus, \( n_{\text{benzene}} = \frac{m \times 92}{7176} \) and \( n_{\text{toluene}} = \frac{m \times 78}{7176} \)
The mole fractions are:
- Mole fraction of benzene, \( X_{\text{benzene}} = \frac{\frac{m \times 92}{7176}}{\frac{m \times 92}{7176} + \frac{m \times 78}{7176}} = \frac{92}{92 + 78} = \frac{92}{170} \)
- Mole fraction of toluene, \( X_{\text{toluene}} = \frac{78}{170} \)
Now, applying Raoult's Law for solution B:
- Partial pressure of benzene, \( P_{\text{benzene}} = X_{\text{benzene}} \times P^{\circ}_{\text{benzene}} = \frac{92}{170} \times 160 \approx 87.76 \, \text{mm Hg} \)
- Partial pressure of toluene, \( P_{\text{toluene}} = X_{\text{toluene}} \times P^{\circ}_{\text{toluene}} = \frac{78}{170} \times 60 \approx 27.53 \, \text{mm Hg} \)
The total pressure \( P_B \) for solution B is:
- \( P_B = P_{\text{benzene}} + P_{\text{toluene}} \approx 87.76 + 27.53 \approx 115.29 \, \text{mm Hg} \)
3. Final Calculation
Now, we can find the sum of the total pressures \( P_A + P_B \):
- \( P_A + P_B = 110 + 115.29 \approx 225.29 \, \text{mm Hg} \)
Rounding off the answer gives us:
- Final answer: \( P_A + P_B \approx 225 \, \text{mm Hg} \)
In summary, after applying Raoult's Law to both solutions, we found that the total pressures sum up to approximately 225 mm Hg. This exercise illustrates the importance of understanding mole fractions and how they relate to vapor pressures in ideal solutions.
