Benzene and toluene form nearly ideal solution. At a certain temperature, the vapour pressure of the pure benzene is 150 mm Hg and of pure toluene is 50 mm Hg. For this temperature, calculate the vapour pressure of solution containing equal weights of two substances. Also calculate their composition in the vapour phase.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the moles of benzene and toluene
Given that we have equal weights of benzene and toluene, we can denote the weight of each as \( W \).

- Molar mass of benzene (C₆H₆) = 78 g/mol
- Molar mass of toluene (C₇H₈) = 92 g/mol

The number of moles of benzene (\( n_B \)) and toluene (\( n_T \)) can be calculated using the formula:

\[ n_B = \frac{W}{M_B} \quad \text{and} \quad n_T = \frac{W}{M_T} \]

Where:
- \( M_B \) = molar mass of benzene = 78 g/mol
- \( M_T \) = molar mass of toluene = 92 g/mol

Since we have equal weights of both substances, we can express the moles as:

\[ n_B = \frac{W}{78} \quad \text{and} \quad n_T = \frac{W}{92} \]

### Step 2: Calculate the total number of moles
The total number of moles (\( n_{total} \)) is given by:

\[ n_{total} = n_B + n_T = \frac{W}{78} + \frac{W}{92} \]

To combine these fractions, we find a common denominator (which is 78 * 92):

\[ n_{total} = \frac{W \cdot 92 + W \cdot 78}{78 \cdot 92} = \frac{W(92 + 78)}{78 \cdot 92} = \frac{W \cdot 170}{78 \cdot 92} \]

### Step 3: Calculate the mole fractions
Now, we can calculate the mole fractions of benzene (\( x_B \)) and toluene (\( x_T \)):

\[ x_B = \frac{n_B}{n_{total}} = \frac{\frac{W}{78}}{\frac{W \cdot 170}{78 \cdot 92}} = \frac{92}{170} \approx 0.541 \]

\[ x_T = \frac{n_T}{n_{total}} = \frac{\frac{W}{92}}{\frac{W \cdot 170}{78 \cdot 92}} = \frac{78}{170} \approx 0.459 \]

### Step 4: Calculate the total vapor pressure of the solution
Using Raoult's Law, the total vapor pressure (\( P_T \)) of the solution is given by:

\[ P_T = P_{B}^0 \cdot x_B + P_{T}^0 \cdot x_T \]

Where:
- \( P_{B}^0 \) = vapor pressure of pure benzene = 150 mm Hg
- \( P_{T}^0 \) = vapor pressure of pure toluene = 50 mm Hg

Substituting the values:

\[ P_T = (150 \, \text{mm Hg} \cdot 0.541) + (50 \, \text{mm Hg} \cdot 0.459) \]

Calculating this gives:

\[ P_T = 81.15 + 22.95 = 104 \, \text{mm Hg} \]

### Step 5: Calculate the composition in the vapor phase
To find the composition in the vapor phase, we can use the following equations:

For benzene in the vapor phase (\( y_B \)):

\[ y_B = \frac{P_B}{P_T} = \frac{P_{B}^0 \cdot x_B}{P_T} \]

Substituting the values:

\[ y_B = \frac{150 \cdot 0.541}{104} \approx 0.778 \]

For toluene in the vapor phase (\( y_T \)):

\[ y_T = \frac{P_T - P_B}{P_T} = 1 - y_B \approx 0.222 \]

### Final Answers
- Total vapor pressure of the solution: **104 mm Hg**
- Composition in the vapor phase:
- Benzene (\( y_B \)): **0.778**
- Toluene (\( y_T \)): **0.222**