Total Vapour pressure of mixture of 1molA`(p_(A)^(0)=150 "torr")`and 2molB `(p_(B)^(0)=240 "torr")`is `200 "torr"`. In this case

Options

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

Explanation

To solve the problem, we need to calculate the total vapor pressure of a mixture of two components, A and B, using Raoult's law. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Moles of A (n_A) = 1 mol
- Moles of B (n_B) = 2 mol
- Vapor pressure of pure A (P_A^0) = 150 torr
- Vapor pressure of pure B (P_B^0) = 240 torr
- Total vapor pressure (P_total) = 200 torr

### Step 2: Calculate the total number of moles in the mixture
Total moles (n_total) = n_A + n_B = 1 + 2 = 3 moles

### Step 3: Calculate the mole fractions of A and B
- Mole fraction of A (X_A) = n_A / n_total = 1 / 3
- Mole fraction of B (X_B) = n_B / n_total = 2 / 3

### Step 4: Apply Raoult's Law to find the partial pressures
Using Raoult's law:
- Partial pressure of A (P_A) = X_A * P_A^0 = (1/3) * 150 torr = 50 torr
- Partial pressure of B (P_B) = X_B * P_B^0 = (2/3) * 240 torr = 160 torr

### Step 5: Calculate the total vapor pressure
Total vapor pressure (P_total) = P_A + P_B = 50 torr + 160 torr = 210 torr

### Step 6: Compare with the given total vapor pressure
The calculated total vapor pressure (210 torr) is greater than the given total vapor pressure (200 torr). This indicates that the solution exhibits a negative deviation from Raoult's law.

### Conclusion
The mixture shows a negative deviation from ideal behavior, as the observed total vapor pressure is less than what would be expected from the ideal behavior predicted by Raoult's law.

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