Liquids A`(p_(A)^(@)=360" mm Hg")` and `B(p_(B)^(@)=320" mm Hg")` are mixed. If solution has vapour has vapour pressure 340 mm Hg, then number of mole fraction of B/mole solution will be

Options

A. `(33)/(34)`
B. `(1)/(34)`
C. `(1)/(2)`
D. `(1)/(4)`

Answer

To solve the problem step by step, we will use Raoult's Law and the concept of mole fractions.

### Step 1: Understand Raoult's Law
Raoult's Law states that the partial pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapor pressure in the pure state.

### Step 2: Define Variables
Let:
- \( P_A^0 = 360 \, \text{mm Hg} \) (vapor pressure of pure A)
- \( P_B^0 = 320 \, \text{mm Hg} \) (vapor pressure of pure B)
- \( P_{\text{total}} = 340 \, \text{mm Hg} \) (vapor pressure of the solution)
- \( x_A \) = mole fraction of A
- \( x_B \) = mole fraction of B

From the definition of mole fractions, we know that:
\[ x_A + x_B = 1 \]
This implies:
\[ x_A = 1 - x_B \]

### Step 3: Apply Raoult's Law
According to Raoult's Law, the total vapor pressure of the solution can be expressed as:
\[ P_{\text{total}} = x_A P_A^0 + x_B P_B^0 \]

Substituting the known values:
\[ 340 = x_A \cdot 360 + x_B \cdot 320 \]

### Step 4: Substitute \( x_A \)
Now, substitute \( x_A \) with \( 1 - x_B \):
\[ 340 = (1 - x_B) \cdot 360 + x_B \cdot 320 \]

### Step 5: Expand and Rearrange
Expanding the equation:
\[ 340 = 360 - 360x_B + 320x_B \]
Combining like terms:
\[ 340 = 360 - 40x_B \]

### Step 6: Solve for \( x_B \)
Rearranging the equation to isolate \( x_B \):
\[ 40x_B = 360 - 340 \]
\[ 40x_B = 20 \]
\[ x_B = \frac{20}{40} = \frac{1}{2} \]

### Step 7: Conclusion
The mole fraction of B in the solution is:
\[ x_B = \frac{1}{2} \]