Question

38.4 g of an unknown substance (molar mass \(= 384\ \text{g mol}^{-1}\)) and 116 g of acetone are used to prepare a solution at 313 K. If vapour pressure of pure acetone (molar mass \(= 58\ \text{g mol}^{-1}\)) is 0.842 atm, what is the vapour pressure of the solution?

• A. 0.650 atm
• B. 0.880 atm
• C. 0.7999 atm
• D. 0.958 atm

Hint:

For a non-volatile solute, vapour pressure depends only on the mole fraction of the solvent.

Answer 1

The Correct Option is C

Step 1: Calculate number of moles of components.
Moles of unknown substance: \[ \frac{38.4}{384} = 0.1\ \text{mol} \]
Moles of acetone: \[ \frac{116}{58} = 2\ \text{mol} \]

Step 2: Calculate mole fraction of acetone.
Total moles \(= 0.1 + 2 = 2.1\)
\[ X_{\text{acetone}} = \frac{2}{2.1} = 0.9524 \]

Step 3: Apply Raoult’s law.
\[ P_{\text{solution}} = X_{\text{acetone}} \times P^\circ_{\text{acetone}} \]
\[ P = 0.9524 \times 0.842 = 0.7999\ \text{atm} \]

Step 4: Conclusion.
The vapour pressure of the solution is \(0.7999\ \text{atm}\).