V.P. of solvent in pure state is 600 mm Hg, when a non-volatile solute is added to it vapour pressure of solution becomes 594 mm Hg, then `x_(B)` will be

Options

A. 0.01
B. 0.1
C. 0.99
D. 0.9

Answer

To find the mole fraction of the solvent (denoted as \( x_B \)) when a non-volatile solute is added, we can use the formula derived from Raoult's Law, which relates the vapor pressure of the solvent in pure state (\( P_0 \)) and the vapor pressure of the solution (\( P_s \)).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of the solvent in pure state, \( P_0 = 600 \, \text{mm Hg} \)
- Vapor pressure of the solution after adding the non-volatile solute, \( P_s = 594 \, \text{mm Hg} \)

2. **Use the Formula:**
The formula to calculate the mole fraction of the solvent (\( x_B \)) is:
\[ x_B = \frac{P_0 - P_s}{P_0} \]

3. **Substitute the Values:**
Substitute the values of \( P_0 \) and \( P_s \) into the formula:
\[ x_B = \frac{600 \, \text{mm Hg} - 594 \, \text{mm Hg}}{600 \, \text{mm Hg}} \]

4. **Calculate the Numerator:**
Calculate the difference in vapor pressures:
\[ 600 - 594 = 6 \, \text{mm Hg} \]

5. **Complete the Calculation:**
Now substitute this value back into the equation:
\[ x_B = \frac{6 \, \text{mm Hg}}{600 \, \text{mm Hg}} = 0.01 \]

6. **Final Result:**
The mole fraction of the solvent (\( x_B \)) is:
\[ x_B = 0.01 \]

### Conclusion:
The mole fraction of the solvent in the solution is \( 0.01 \).