1 mol benzene `(P^(@)_("benzene")=42 mm)` and 2 mol toluence `(P^(@)_("toluene")=36 mm)` will have

Options

A. Total vapour pressure of `38 mm`.
B. Mole fraction of vapour of benzene above liquid mixture is `7//19`.
C. Positive devaition from Raoult's law.
D. Negative devaition from Raoult's law.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Moles of benzene (n_benzene) = 1 mol
- Moles of toluene (n_toluene) = 2 mol
- Vapor pressure of pure benzene (P^0_benzene) = 42 mm Hg
- Vapor pressure of pure toluene (P^0_toluene) = 36 mm Hg

### Step 2: Calculate the total number of moles
Total moles (n_total) = n_benzene + n_toluene = 1 mol + 2 mol = 3 mol

### Step 3: Calculate the mole fractions of benzene and toluene
- Mole fraction of benzene (x_benzene) = n_benzene / n_total = 1 mol / 3 mol = 1/3
- Mole fraction of toluene (x_toluene) = n_toluene / n_total = 2 mol / 3 mol = 2/3

### Step 4: Apply Raoult's Law to find the total vapor pressure
According to Raoult's Law:
\[ P_{total} = P^0_{benzene} \cdot x_{benzene} + P^0_{toluene} \cdot x_{toluene} \]

Substituting the values:
\[ P_{total} = (42 \, \text{mm Hg}) \cdot \left(\frac{1}{3}\right) + (36 \, \text{mm Hg}) \cdot \left(\frac{2}{3}\right) \]

Calculating each term:
- For benzene: \( 42 \cdot \frac{1}{3} = 14 \, \text{mm Hg} \)
- For toluene: \( 36 \cdot \frac{2}{3} = 24 \, \text{mm Hg} \)

Now, add these two contributions:
\[ P_{total} = 14 \, \text{mm Hg} + 24 \, \text{mm Hg} = 38 \, \text{mm Hg} \]

### Step 5: Calculate the mole fraction of benzene in the vapor phase
Using the formula:
\[ y_{benzene} = \frac{P_{benzene}}{P_{total}} \]
Where \( P_{benzene} = P^0_{benzene} \cdot x_{benzene} \)

We already calculated \( P_{benzene} \):
\[ P_{benzene} = 14 \, \text{mm Hg} \]

Now substituting into the equation:
\[ y_{benzene} = \frac{14 \, \text{mm Hg}}{38 \, \text{mm Hg}} \]

Calculating:
\[ y_{benzene} = \frac{14}{38} = \frac{7}{19} \]

### Final Answer
The total vapor pressure of the mixture is **38 mm Hg**, and the mole fraction of benzene in the vapor phase is **7/19**.

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