Question

Vapour pressure of a pure liquid solvent A is $0.80\,\mathrm{atm}$. When a non-volatile solute B is added to the solvent, its vapour pressure drops to $0.60\,\mathrm{atm}$. The mole fraction of components A and B in the solution are respectively:}

• A. 0.75, 0.25
• B. 0.25, 0.75
• C. 0.60, 0.40
• D. 0.40, 0.60

Hint:

For a non-volatile solute: \[ X_{\text{solvent}} = \frac{P_{\text{solution}}}{P^\circ_{\text{solvent}}} \] Then use \[ X_{\text{solute}}=1-X_{\text{solvent}}. \]

Answer 1

The Correct Option is A

Concept: For a solution containing a non-volatile solute, Raoult's law states that: \[ P=P^\circ X_A \] where \[ P^\circ=\text{vapour pressure of pure solvent} \] and \[ X_A=\text{mole fraction of solvent}. \]

Step 1: Write the given data. \[ P^\circ =0.80\ \mathrm{atm} \] \[ P=0.60\ \mathrm{atm} \]

Step 2: Apply Raoult's law. \[ P=P^\circ X_A \] Substituting values: \[ 0.60=0.80X_A \] \[ X_A=\frac{0.60}{0.80} \] \[ X_A=0.75 \]

Step 3: Calculate mole fraction of solute. Since \[ X_A+X_B=1 \] therefore \[ X_B=1-0.75 \] \[ X_B=0.25 \]

Step 4: Write the answer. \[ \boxed{X_A=0.75,\qquad X_B=0.25} \] Hence, \[ \boxed{\text{Option (1)}} \]