Vapour pressure of pure A = 100 torr, moles = 2. Vapour pressure of pure B = 80 torr, moles = 3. Total vapour pressure of mixture is

Options

A. 440 torr
B. 460 torr
C. 180 torr
D. 88 torr

Explanation

To solve the problem of finding the total vapor pressure of a mixture of two components A and B using Raoult's Law, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Vapor pressure of pure A, \( P^0_A = 100 \) torr
- Moles of A, \( n_A = 2 \)
- Vapor pressure of pure B, \( P^0_B = 80 \) torr
- Moles of B, \( n_B = 3 \)

2. **Calculate Total Moles:**
\[ n_{total} = n_A + n_B = 2 + 3 = 5 \text{ moles} \]

3. **Calculate Mole Fractions:**
- Mole fraction of A, \( X_A \):
\[ X_A = \frac{n_A}{n_{total}} = \frac{2}{5} \]
- Mole fraction of B, \( X_B \):
\[ X_B = \frac{n_B}{n_{total}} = \frac{3}{5} \]

4. **Apply Raoult's Law:**
According to Raoult's Law, the total vapor pressure \( P_t \) of the mixture is given by:
\[ P_t = P^0_A \cdot X_A + P^0_B \cdot X_B \]

5. **Substitute Values:**
\[ P_t = (100 \, \text{torr}) \cdot \left(\frac{2}{5}\right) + (80 \, \text{torr}) \cdot \left(\frac{3}{5}\right) \]

6. **Calculate Each Component:**
- For A:
\[ P^0_A \cdot X_A = 100 \cdot \frac{2}{5} = 40 \, \text{torr} \]
- For B:
\[ P^0_B \cdot X_B = 80 \cdot \frac{3}{5} = 48 \, \text{torr} \]

7. **Add the Contributions:**
\[ P_t = 40 \, \text{torr} + 48 \, \text{torr} = 88 \, \text{torr} \]

### Final Answer:
The total vapor pressure of the mixture is **88 torr**.

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