On mixing 1 mole of `C_(6)H_(6)(P^(@)=42mm)` and 2 mole of `C_(7)H_(8)(P^(@)=36mm)` one can conclude

Options

A. total vapour pressure of mixture =38mm
B. mole fraction of vapours of `C_(6)H_(6)` above liquid mixture `=(7)/(19)`
C. positive deviation from Raoult's law
D. ideal solution of two

Answer

To solve the problem of mixing 1 mole of benzene (C₆H₆) and 2 moles of toluene (C₇H₈), we will apply Raoult's Law and calculate the total vapor pressure and the mole fraction of benzene in the mixture.

### Step-by-Step Solution:

**Step 1: Identify the given data.**
- Moles of benzene (C₆H₆) = 1 mole
- Moles of toluene (C₇H₈) = 2 moles
- Vapor pressure of pure benzene (P₀A) = 42 mm Hg
- Vapor pressure of pure toluene (P₀B) = 36 mm Hg

**Hint for Step 1:** Make sure to note the moles and vapor pressures of each component clearly.

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**Step 2: Calculate the total moles in the mixture.**
- Total moles = Moles of benzene + Moles of toluene = 1 + 2 = 3 moles

**Hint for Step 2:** Always sum the moles of all components to find the total moles in the solution.

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**Step 3: Calculate the mole fractions of benzene and toluene.**
- Mole fraction of benzene (X_A) = Moles of benzene / Total moles = 1 / 3
- Mole fraction of toluene (X_B) = Moles of toluene / Total moles = 2 / 3

**Hint for Step 3:** The mole fraction is calculated by dividing the number of moles of each component by the total number of moles.

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**Step 4: Apply Raoult's Law to find the partial pressures.**
- Partial pressure of benzene (P_A) = P₀A * X_A = 42 mm Hg * (1/3) = 14 mm Hg
- Partial pressure of toluene (P_B) = P₀B * X_B = 36 mm Hg * (2/3) = 24 mm Hg

**Hint for Step 4:** Raoult's Law states that the partial pressure of each component is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

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**Step 5: Calculate the total vapor pressure (P_total).**
- P_total = P_A + P_B = 14 mm Hg + 24 mm Hg = 38 mm Hg

**Hint for Step 5:** The total vapor pressure is the sum of the partial pressures of all components in the mixture.

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**Step 6: Calculate the mole fraction of benzene in the vapor phase.**
- Mole fraction of benzene in vapor (Y_A) = P_A / P_total = 14 mm Hg / 38 mm Hg = 7/19

**Hint for Step 6:** The mole fraction in the vapor phase is calculated by dividing the partial pressure of the component by the total vapor pressure.

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**Conclusion:**
From the calculations, we have:
- Total vapor pressure = 38 mm Hg
- Mole fraction of benzene in the vapor = 7/19
- The mixture follows Raoult's law, indicating no deviation from ideal behavior.

### Final Answer:
The conclusions drawn from the mixing of 1 mole of benzene and 2 moles of toluene are:
1. Total vapor pressure = 38 mm Hg
2. Mole fraction of benzene in the vapor = 7/19
3. The solution behaves ideally, following Raoult's law.