Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively

Options

A. 200 and 300
B. 300 and 400
C. 400 and 600
D. 500 and 600

Answer

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial pressures of its components. Let's break down the solution step by step.

### Step 1: Identify the initial conditions
We have:
- 1 mole of liquid X
- 3 moles of liquid Y
- Total moles = 1 + 3 = 4 moles
- Vapor pressure of the solution = 550 mm Hg

### Step 2: Calculate the mole fractions
The mole fraction of X (X_X) and Y (X_Y) can be calculated as follows:
- Mole fraction of X, \( X_X = \frac{1}{4} \)
- Mole fraction of Y, \( X_Y = \frac{3}{4} \)

### Step 3: Apply Raoult's Law
According to Raoult's Law:
\[ P_{solution} = X_X \cdot P^0_X + X_Y \cdot P^0_Y \]
Substituting the known values:
\[ 550 = \left(\frac{1}{4}\right) P^0_X + \left(\frac{3}{4}\right) P^0_Y \tag{1} \]

### Step 4: Consider the new condition after adding 1 mole of Y
When 1 mole of Y is added, the new composition becomes:
- 1 mole of X
- 4 moles of Y
- Total moles = 1 + 4 = 5 moles

Now, the new mole fractions are:
- Mole fraction of X, \( X_X = \frac{1}{5} \)
- Mole fraction of Y, \( X_Y = \frac{4}{5} \)

### Step 5: New vapor pressure condition
The new vapor pressure of the solution increases by 10 mm Hg, so:
\[ P_{new} = 550 + 10 = 560 \text{ mm Hg} \]
Using Raoult's Law again:
\[ 560 = \left(\frac{1}{5}\right) P^0_X + \left(\frac{4}{5}\right) P^0_Y \tag{2} \]

### Step 6: Solve the system of equations
Now we have two equations:
1. \( \frac{1}{4} P^0_X + \frac{3}{4} P^0_Y = 550 \)
2. \( \frac{1}{5} P^0_X + \frac{4}{5} P^0_Y = 560 \)

**Multiply equation (1) by 4:**
\[ P^0_X + 3 P^0_Y = 2200 \tag{3} \]

**Multiply equation (2) by 5:**
\[ P^0_X + 4 P^0_Y = 2800 \tag{4} \]

### Step 7: Subtract equation (3) from equation (4)
\[ (P^0_X + 4 P^0_Y) - (P^0_X + 3 P^0_Y) = 2800 - 2200 \]
This simplifies to:
\[ P^0_Y = 600 \text{ mm Hg} \]

### Step 8: Substitute \( P^0_Y \) back into equation (3)
\[ P^0_X + 3(600) = 2200 \]
\[ P^0_X + 1800 = 2200 \]
\[ P^0_X = 400 \text{ mm Hg} \]

### Final Answer
The vapor pressures of X and Y in their pure states are:
- \( P^0_X = 400 \text{ mm Hg} \)
- \( P^0_Y = 600 \text{ mm Hg} \)