If `P^(@)` the vapour pressure of a pure solvent and P is the vapour pressure of the solution prepared by dissolving a non volatile solute in it. The mole fraction of the solvent `X_(A)` is given by:

Options

A. `(P^(@) -P)/(P^(@)) = X_(A)`
B. `(P^(@)-P)/(P) =X_(A)`
C. `(P)/(P^(@)) = X_(A)`
D. `P^(@) - P = X_(A)`

Explanation

To find the mole fraction of the solvent \( X_A \) in a solution prepared by dissolving a non-volatile solute, we can use the concept of relative lowering of vapor pressure. Here’s a step-by-step solution:

### Step 1: Understand the Concept of Vapor Pressure
The vapor pressure of a pure solvent is denoted as \( P^0 \) (or \( P_0 \)), and the vapor pressure of the solution containing a non-volatile solute is denoted as \( P \).

### Step 2: Apply the Formula for Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure can be expressed as:
\[ \text{Relative lowering of vapor pressure} = \frac{P^0 - P}{P^0} \]
This relative lowering is equal to the mole fraction of the solute (\( X_B \)) in the solution.

### Step 3: Relate Mole Fractions
Since the mole fraction of the solute \( X_B \) and the mole fraction of the solvent \( X_A \) are related by:
\[ X_A + X_B = 1 \]
we can express \( X_B \) as:
\[ X_B = 1 - X_A \]

### Step 4: Substitute into the Relative Lowering Equation
Substituting \( X_B \) into the relative lowering formula gives us:
\[ \frac{P^0 - P}{P^0} = X_B = 1 - X_A \]

### Step 5: Rearranging the Equation
Now, rearranging the equation to find \( X_A \):
\[ X_A = 1 - \frac{P^0 - P}{P^0} \]
This simplifies to:
\[ X_A = \frac{P}{P^0} \]

### Final Result
Thus, the mole fraction of the solvent \( X_A \) is given by:
\[ X_A = \frac{P}{P^0} \]