Two volatile liquide A and B are mixed in mole ratio 1:3, to form an ideal liquid solution Vapour pressure of pure. A and B are 600 torr and 200 torr. Volume of container for vapour is 76 litre. `(R= 1//12" " "litre" " "atm//K//"mole")`
Calculate the mole of A in vapour phase at 300 K, assuming negligible amount of vapour is present compared to liquid in container

Options

A. `2` mole
B. `0.3` mole
C. `1.2` mole
D. `0.6` mole

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the moles of A and B in the solution
Given the mole ratio of A to B is 1:3, we can assume:
- Moles of A = 1 mole
- Moles of B = 3 moles

### Step 2: Calculate the mole fractions of A and B
The total number of moles in the solution is:
\[ \text{Total moles} = \text{Moles of A} + \text{Moles of B} = 1 + 3 = 4 \text{ moles} \]

Now, we can calculate the mole fractions:
- Mole fraction of A (\(X_A\)):
\[ X_A = \frac{\text{Moles of A}}{\text{Total moles}} = \frac{1}{4} = 0.25 \]

- Mole fraction of B (\(X_B\)):
\[ X_B = \frac{\text{Moles of B}}{\text{Total moles}} = \frac{3}{4} = 0.75 \]

### Step 3: Calculate the partial vapor pressures using Raoult's Law
The vapor pressures of pure A and B are given as:
- Vapor pressure of pure A (\(P^0_A\)) = 600 torr
- Vapor pressure of pure B (\(P^0_B\)) = 200 torr

Using Raoult's Law:
- Partial pressure of A (\(P_A\)):
\[ P_A = X_A \times P^0_A = 0.25 \times 600 \text{ torr} = 150 \text{ torr} \]

- Partial pressure of B (\(P_B\)):
\[ P_B = X_B \times P^0_B = 0.75 \times 200 \text{ torr} = 150 \text{ torr} \]

### Step 4: Calculate the total vapor pressure of the solution
The total vapor pressure (\(P_{total}\)) is the sum of the partial pressures:
\[ P_{total} = P_A + P_B = 150 \text{ torr} + 150 \text{ torr} = 300 \text{ torr} \]

### Step 5: Convert total vapor pressure to atm
To convert torr to atm:
\[ 1 \text{ atm} = 760 \text{ torr} \]
Thus,
\[ P_{total} = \frac{300 \text{ torr}}{760 \text{ torr/atm}} = 0.3947 \text{ atm} \]

### Step 6: Use the ideal gas law to find the total number of moles in the vapor phase
Using the ideal gas law:
\[ PV = nRT \]
Where:
- \(P = 0.3947 \text{ atm}\)
- \(V = 76 \text{ L}\)
- \(R = 0.0821 \text{ L atm/(K mol)}\)
- \(T = 300 \text{ K}\)

Rearranging for \(n\):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(0.3947 \text{ atm}) \times (76 \text{ L})}{(0.0821 \text{ L atm/(K mol)}) \times (300 \text{ K})} \]

Calculating:
\[ n = \frac{29.97}{24.63} \approx 1.22 \text{ moles} \]

### Step 7: Calculate the moles of A in the vapor phase
The moles of A in the vapor phase can be calculated using its mole fraction:
\[ \text{Moles of A in vapor} = X_A \times n = 0.25 \times 1.22 \approx 0.305 \text{ moles} \]

### Final Answer
The moles of A in the vapor phase at 300 K is approximately **0.305 moles**.

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