Total vapour pressure of mixture of 1 mole of volatile component A `(P_(A)^(@)=100 mm Hg)` and 3 mole of volatile component `B (P_(B)^(@)=80 mm Hg)` is `90 mm Hg`. For such case `:`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- For component A:
- Number of moles (nA) = 1 mole
- Vapor pressure at pure state (P₀A) = 100 mm Hg
- For component B:
- Number of moles (nB) = 3 moles
- Vapor pressure at pure state (P₀B) = 80 mm Hg
- Total vapor pressure of the mixture (P_total) = 90 mm Hg

### Step 2: Calculate the total number of moles in the mixture
Total moles (n_total) = nA + nB = 1 + 3 = 4 moles

### Step 3: Calculate the mole fractions of components A and B
- Mole fraction of A (xA) = nA / n_total = 1 / 4 = 0.25
- Mole fraction of B (xB) = nB / n_total = 3 / 4 = 0.75

### Step 4: Calculate the expected total vapor pressure using Raoult's Law
According to Raoult's Law, the total vapor pressure (P_total) of the solution can be calculated as:
\[ P_{total} = P_{0A} \cdot x_A + P_{0B} \cdot x_B \]

Substituting the values:
\[ P_{total} = (100 \, \text{mm Hg} \cdot 0.25) + (80 \, \text{mm Hg} \cdot 0.75) \]

### Step 5: Perform the calculations
1. Calculate the contribution from A:
\[ 100 \, \text{mm Hg} \cdot 0.25 = 25 \, \text{mm Hg} \]

2. Calculate the contribution from B:
\[ 80 \, \text{mm Hg} \cdot 0.75 = 60 \, \text{mm Hg} \]

3. Add the contributions:
\[ P_{total} = 25 \, \text{mm Hg} + 60 \, \text{mm Hg} = 85 \, \text{mm Hg} \]

### Step 6: Compare the calculated total vapor pressure with the given total vapor pressure
The calculated total vapor pressure (85 mm Hg) is less than the given total vapor pressure (90 mm Hg). This indicates that the mixture exhibits a positive deviation from Raoult's Law.

### Conclusion
The final conclusion is that the mixture shows a positive deviation from Raoult's Law because the observed vapor pressure (90 mm Hg) is greater than the calculated vapor pressure (85 mm Hg).

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