The total vapour pressure of a mixture of 1 mole of A (`P_A = 200 mm Hg`) and 3 mole of B (`P_B= 360` mm Hg) is 350 mm Hg. Then

Options

A. There is no deviation from Raoult's law
B. There is positive deviation from Raoult's law
C. There is negative deviation from Raoult's law
D. Molecular masses of A and B are also required

Answer

To solve the problem, we need to determine the nature of the deviation from Raoult's law for the given mixture of substances A and B.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Moles of A (n_A) = 1 mole
- Moles of B (n_B) = 3 moles
- Vapor pressure of pure A (P^0_A) = 200 mm Hg
- Vapor pressure of pure B (P^0_B) = 360 mm Hg
- Total vapor pressure (P_total) = 350 mm Hg

2. **Calculate the Mole Fractions:**
- Total moles in the mixture (n_total) = n_A + n_B = 1 + 3 = 4 moles
- Mole fraction of A (x_A) = n_A / n_total = 1 / 4 = 0.25
- Mole fraction of B (x_B) = n_B / n_total = 3 / 4 = 0.75

3. **Calculate the Expected Vapor Pressure Using Raoult's Law:**
- Expected vapor pressure of A (P_A) = P^0_A * x_A = 200 mm Hg * 0.25 = 50 mm Hg
- Expected vapor pressure of B (P_B) = P^0_B * x_B = 360 mm Hg * 0.75 = 270 mm Hg

4. **Calculate the Total Expected Vapor Pressure:**
- Total expected vapor pressure (P_expected) = P_A + P_B = 50 mm Hg + 270 mm Hg = 320 mm Hg

5. **Compare the Total Expected Vapor Pressure with the Given Total Vapor Pressure:**
- Given total vapor pressure = 350 mm Hg
- Expected total vapor pressure = 320 mm Hg

6. **Determine the Nature of Deviation:**
- Since the given total vapor pressure (350 mm Hg) is greater than the expected total vapor pressure (320 mm Hg), this indicates a positive deviation from Raoult's law.

7. **Conclusion:**
- The correct option is that there is a positive deviation from Raoult's law.