The vapour pressure of pure `CHCl_(3) and CH_(2)Cl_(2)` are 200 ad 41.5 atm respectively. The weight of `CHCl_(3) and CH_(2)Cl_(2)` are respectively 11.9 g and 17 gm. The vapour pressure of solution will be

Options

A. 80.5
B. 79.5
C. 94.3
D. 105.5

Explanation

To solve the problem of finding the vapor pressure of the solution containing CHCl3 and CH2Cl2, we can follow these steps:

### Step 1: Calculate the number of moles of each component.

1. **For CHCl3:**
- Given weight = 11.9 g
- Molar mass of CHCl3 = 119 g/mol
- Number of moles of CHCl3 = Weight / Molar mass = 11.9 g / 119 g/mol = 0.1 moles

2. **For CH2Cl2:**
- Given weight = 17 g
- Molar mass of CH2Cl2 = 85 g/mol
- Number of moles of CH2Cl2 = Weight / Molar mass = 17 g / 85 g/mol = 0.2 moles

### Step 2: Calculate the mole fractions of each component.

1. **Mole fraction of CHCl3 (χ_CHCl3):**
- Total moles = moles of CHCl3 + moles of CH2Cl2 = 0.1 + 0.2 = 0.3
- χ_CHCl3 = moles of CHCl3 / total moles = 0.1 / 0.3 = 1/3

2. **Mole fraction of CH2Cl2 (χ_CH2Cl2):**
- χ_CH2Cl2 = moles of CH2Cl2 / total moles = 0.2 / 0.3 = 2/3

### Step 3: Calculate the vapor pressure of the solution.

Using Raoult's Law:
- Vapor pressure of solution (P_solution) = (P°_CHCl3 * χ_CHCl3) + (P°_CH2Cl2 * χ_CH2Cl2)

Where:
- P°_CHCl3 = 200 atm (vapor pressure of pure CHCl3)
- P°_CH2Cl2 = 41.5 atm (vapor pressure of pure CH2Cl2)

Substituting the values:
- P_solution = (200 atm * (1/3)) + (41.5 atm * (2/3))

Calculating each term:
- P_solution = (200/3) + (41.5 * 2/3)
- P_solution = (200/3) + (83/3)
- P_solution = (200 + 83) / 3
- P_solution = 283 / 3
- P_solution ≈ 94.33 atm

### Final Answer:
The vapor pressure of the solution is approximately **94.33 atm**.

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