Vapour pressure of pure chloroform `(CHCl_(3))` and dichloromethane `(CH_(2)Cl_(2))` at `25^(@)C` are 200 mmHg and 41.5 mmHg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of `CHCl_(3)` and 40 g of `CH_(2)Cl_(2)` at the same temperature will be :
(Molecular mass of `CHCl_(3)` = 119.5 u and molecular mass of `CH_(2)Cl_(2) = 85u`)

Options

A. 173.9 mmHg
B. 615.0 mmHg
C. 347.9 mmHg
D. 90.63 mmHg.

Answer

To find the vapor pressure of the solution obtained by mixing chloroform (CHCl₃) and dichloromethane (CH₂Cl₂), we will use Raoult's Law. The steps to solve this problem are as follows:

### Step 1: Calculate the number of moles of each component
1. **Moles of CHCl₃ (A)**:
\[ \text{Moles of CHCl₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{25.5 \, \text{g}}{119.5 \, \text{g/mol}} \approx 0.2133 \, \text{mol} \]

2. **Moles of CH₂Cl₂ (B)**:
\[ \text{Moles of CH₂Cl₂} = \frac{40 \, \text{g}}{85 \, \text{g/mol}} \approx 0.4705 \, \text{mol} \]

### Step 2: Calculate the total number of moles
\[ \text{Total moles} = \text{Moles of CHCl₃} + \text{Moles of CH₂Cl₂} = 0.2133 + 0.4705 \approx 0.6838 \, \text{mol} \]

### Step 3: Calculate the mole fractions of each component
1. **Mole fraction of CHCl₃ (Xₐ)**:
\[ Xₐ = \frac{\text{Moles of CHCl₃}}{\text{Total moles}} = \frac{0.2133}{0.6838} \approx 0.3119 \]

2. **Mole fraction of CH₂Cl₂ (Xᵦ)**:
\[ Xᵦ = \frac{\text{Moles of CH₂Cl₂}}{\text{Total moles}} = \frac{0.4705}{0.6838} \approx 0.6880 \]

### Step 4: Calculate the partial pressures using Raoult's Law
1. **Partial pressure of CHCl₃ (Pₐ)**:
\[ Pₐ = P^0ₐ \cdot Xₐ = 200 \, \text{mmHg} \cdot 0.3119 \approx 62.38 \, \text{mmHg} \]

2. **Partial pressure of CH₂Cl₂ (Pᵦ)**:
\[ Pᵦ = P^0ᵦ \cdot Xᵦ = 41.5 \, \text{mmHg} \cdot 0.6880 \approx 28.55 \, \text{mmHg} \]

### Step 5: Calculate the total vapor pressure of the solution
\[ P_{\text{total}} = Pₐ + Pᵦ = 62.38 \, \text{mmHg} + 28.55 \, \text{mmHg} \approx 90.93 \, \text{mmHg} \]

### Final Answer
The vapor pressure of the solution obtained by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 25°C is approximately **90.93 mmHg**.

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