Two liquids A and B form an ideal solution of 1 mole of A and x moles of B is 550 mm of Hg. If the vapour pressures of pure A and B are 400 mm of Hg nd 600 mm of Hg respectively. Then x is-

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem step by step, we will use Raoult's Law and Dalton's Law of Partial Pressures.

### Step 1: Understand the Given Information
We have:
- 1 mole of liquid A.
- x moles of liquid B.
- Total vapor pressure of the solution (P_total) = 550 mmHg.
- Vapor pressure of pure A (P0A) = 400 mmHg.
- Vapor pressure of pure B (P0B) = 600 mmHg.

### Step 2: Apply Dalton's Law of Partial Pressures
According to Dalton's Law, the total pressure of the solution is the sum of the partial pressures of the components:
\[ P_{total} = P_A + P_B \]

### Step 3: Apply Raoult's Law
According to Raoult's Law, the partial pressure of each component in an ideal solution is given by:
\[ P_A = P_{0A} \cdot x_A \]
\[ P_B = P_{0B} \cdot x_B \]

Where:
- \( x_A \) is the mole fraction of A.
- \( x_B \) is the mole fraction of B.

### Step 4: Calculate the Mole Fractions
The mole fractions are calculated as follows:
- Mole fraction of A, \( x_A = \frac{1}{1 + x} \)
- Mole fraction of B, \( x_B = \frac{x}{1 + x} \)

### Step 5: Substitute into Dalton's Law
Substituting the expressions for \( P_A \) and \( P_B \) into the equation from Dalton's Law:
\[ P_{total} = P_A + P_B \]
\[ 550 = P_{0A} \cdot x_A + P_{0B} \cdot x_B \]
\[ 550 = 400 \cdot \frac{1}{1 + x} + 600 \cdot \frac{x}{1 + x} \]

### Step 6: Simplify the Equation
Combine the terms:
\[ 550 = \frac{400 + 600x}{1 + x} \]

### Step 7: Cross-Multiply to Eliminate the Denominator
Cross-multiplying gives:
\[ 550(1 + x) = 400 + 600x \]

### Step 8: Expand and Rearrange the Equation
Expanding the left side:
\[ 550 + 550x = 400 + 600x \]

### Step 9: Solve for x
Rearranging the equation:
\[ 550 - 400 = 600x - 550x \]
\[ 150 = 50x \]
\[ x = 3 \]

### Final Answer
Thus, the value of x is 3.