A solution is prepared by mixing 8.5g of `CH_(2)Cl_(2)` and 11.95g of `CHCl_(3)`. If vapour pressure of `CH_(2)Cl_(2)` and `CHCl_(3)` at 298 K are 415 and 200 mm Hg respectively, the mole fraction of `CHCl_(3)` in vapour form is :
`("Molar mass of Cl"="35.5 g mol"^(-1))`

Options

A. 0.162
B. 0.675
C. 0.325
D. 0.486

Answer

To solve the problem of finding the mole fraction of CHCl₃ in the vapor form, we will follow these steps:

### Step 1: Calculate the molar mass of CH₂Cl₂ and CHCl₃
- **Molar mass of CH₂Cl₂**:
- Carbon (C): 12 g/mol
- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Chlorine (Cl): 35.5 g/mol × 2 = 71 g/mol
- Total = 12 + 2 + 71 = 85 g/mol

- **Molar mass of CHCl₃**:
- Carbon (C): 12 g/mol
- Hydrogen (H): 1 g/mol
- Chlorine (Cl): 35.5 g/mol × 3 = 106.5 g/mol
- Total = 12 + 1 + 106.5 = 119.5 g/mol

### Step 2: Calculate the number of moles of CH₂Cl₂ and CHCl₃
- **Moles of CH₂Cl₂**:
\[ n_{CH_2Cl_2} = \frac{8.5 \text{ g}}{85 \text{ g/mol}} = 0.1 \text{ moles} \]

- **Moles of CHCl₃**:
\[ n_{CHCl_3} = \frac{11.95 \text{ g}}{119.5 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Calculate the mole fractions of CH₂Cl₂ and CHCl₃
- **Total moles**:
\[ n_{total} = n_{CH_2Cl_2} + n_{CHCl_3} = 0.1 + 0.1 = 0.2 \text{ moles} \]

- **Mole fraction of CHCl₃**:
\[ X_{CHCl_3} = \frac{n_{CHCl_3}}{n_{total}} = \frac{0.1}{0.2} = 0.5 \]

### Step 4: Calculate the total vapor pressure of the solution
- **Using Raoult's Law**:
\[ P_{total} = X_{CHCl_3} \cdot P^0_{CHCl_3} + X_{CH_2Cl_2} \cdot P^0_{CH_2Cl_2} \]
- Where:
- \( P^0_{CHCl_3} = 200 \text{ mm Hg} \)
- \( P^0_{CH_2Cl_2} = 415 \text{ mm Hg} \)
- \( X_{CH_2Cl_2} = 1 - X_{CHCl_3} = 0.5 \)

- **Calculating total pressure**:
\[ P_{total} = (0.5 \cdot 200) + (0.5 \cdot 415) = 100 + 207.5 = 307.5 \text{ mm Hg} \]

### Step 5: Calculate the mole fraction of CHCl₃ in the vapor phase
- **Using partial pressures**:
\[ Y_{CHCl_3} = \frac{P_{CHCl_3}}{P_{total}} = \frac{X_{CHCl_3} \cdot P^0_{CHCl_3}}{P_{total}} \]
- Where:
- \( P_{CHCl_3} = X_{CHCl_3} \cdot P^0_{CHCl_3} = 0.5 \cdot 200 = 100 \text{ mm Hg} \)

- **Calculating mole fraction in vapor**:
\[ Y_{CHCl_3} = \frac{100}{307.5} \approx 0.325 \]

### Final Answer:
The mole fraction of CHCl₃ in the vapor form is approximately **0.325**.

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