Question

At T(K), 0.1 moles of a non-volatile solute was dissolved in 0.9 moles of a volatile solvent. The vapour pressure of pure solvent is 0.9 . What is the vapour pressure (in ) of the solution? 

• A. \( 0.89 \)
• B. \( 0.81 \)
• C. \( 0.79 \)
• D. \( 0.71 \)

Hint:

To find the vapor pressure of a solution, use Raoult's Law: \( P_{\text{solution}} = P_{\text{solvent}}^0 \times X_{\text{solvent}} \). Always check the mole fraction carefully for accuracy.

Answer 1

The Correct Option is B

Step 1: Understanding Raoult's Law 
Raoult's Law states that the vapour pressure of a solution (\( P_{\text{solution}} \)) is given by: \[ P_{\text{solution}} = P_{\text{solvent}}^0 \times X_{\text{solvent}} \] where: - \( P_{\text{solvent}}^0 \) is the vapour pressure of the pure solvent. - \( X_{\text{solvent}} \) is the mole fraction of the solvent. 

Step 2: Calculating the Mole Fraction of the Solvent 
\[ X_{\text{solvent}} = \frac{\text{moles of solvent}}{\text{total moles in solution}} \] Given: - Moles of solute = 0.1 - Moles of solvent = 0.9 - Vapour pressure of pure solvent = 0.9 Total moles in the solution: \[ 0.9 + 0.1 = 1.0 \] Mole fraction of the solvent: \[ X_{\text{solvent}} = \frac{0.9}{1.0} = 0.9 \] 

Step 3: Calculating the Vapour Pressure of the Solution 
Applying Raoult's Law: \[ P_{\text{solution}} = 0.9 \times 0.9 = 0.81 \text{ } \] 

Step 4: Evaluating the Given Options 
- Option (1): Incorrect, as \( 0.89 \) is too high.
- Option (2): Correct, as calculated \( P_{\text{solution}} = 0.81 \) .
- Option (3): Incorrect, as \( 0.79 \) is not accurate.
- Option (4): Incorrect, as \( 0.71 \) is too low.
Thus, the correct answer is

 Option (2).