1 mole of liquid A and 2 moles of liquid B make a solution having a total vapour pressure of 38 torr. The vapour pressures of pure A and pure B are 45 torr and 36 torr respectively. The described solution

Options

A. is an ideal solution
B. shows negative deviation
C. is a minimum boiling azeotrope
D. has volume greater than the sum of individual volumes

Explanation

To solve the problem, we need to analyze the given information and apply Raoult's Law to determine the nature of the solution formed by mixing liquids A and B.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Moles of liquid A (nA) = 1 mole
- Moles of liquid B (nB) = 2 moles
- Total moles (n_total) = nA + nB = 1 + 2 = 3 moles
- Vapor pressure of pure A (P°A) = 45 torr
- Vapor pressure of pure B (P°B) = 36 torr
- Total vapor pressure of the solution (P_total) = 38 torr

2. **Calculate the Mole Fractions:**
- Mole fraction of A (X_A) = nA / n_total = 1 / 3
- Mole fraction of B (X_B) = nB / n_total = 2 / 3

3. **Calculate the Partial Pressures Using Raoult's Law:**
- Partial pressure of A (P_A) = X_A * P°A
- P_A = (1/3) * 45 torr = 15 torr
- Partial pressure of B (P_B) = X_B * P°B
- P_B = (2/3) * 36 torr = 24 torr

4. **Calculate the Expected Total Vapor Pressure:**
- Expected total vapor pressure (P_expected) = P_A + P_B
- P_expected = 15 torr + 24 torr = 39 torr

5. **Compare the Expected and Actual Vapor Pressures:**
- Actual vapor pressure of the solution (P_total) = 38 torr
- Since P_total (38 torr) < P_expected (39 torr), this indicates a negative deviation from Raoult's Law.

6. **Conclusion:**
- The solution formed by mixing 1 mole of liquid A and 2 moles of liquid B shows negative deviation from Raoult's Law.

### Final Answer:
The described solution shows negative deviation.

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