Question

The vapour pressure of pure water is 23.5 mm Hg. Then, the vapour pressure of an aqueous solution which contains 5 mass percent of urea is (Molar mass of urea is 60).

• A. 23 mm Hg
• B. 18 mm Hg
• C. 31 mm Hg
• D. 35 mm Hg

Hint:

For a dilute solution of a non-volatile solute, the change in vapour pressure is small, and the vapour pressure of the solution is close to the pure solvent's vapour pressure.

Answer 1

The Correct Option is A

Raoult’s Law states that the vapour pressure of a solvent in a solution is proportional to the mole fraction of the solvent. This law can be expressed as: \[ P_{\text{solvent}} = P_{\text{solvent}}^0 \times X_{\text{solvent}} \] where \( P_{\text{solvent}} \) is the vapour pressure of the solvent in the solution, \( P_{\text{solvent}}^0 \) is the vapour pressure of the pure solvent, and \( X_{\text{solvent}} \) is the mole fraction of the solvent. Since urea is a non-volatile solute, it does not contribute to the vapour pressure. The presence of the solute reduces the vapour pressure slightly. The change in vapour pressure can be calculated using: \[ \Delta P = P_{\text{solvent}}^0 \times X_{\text{solute}} \] where \( X_{\text{solute}} \) is the mole fraction of the solute. However, because the solution is dilute, the mole fraction of the solute is small, and thus the change in vapour pressure is also small. Therefore, the vapour pressure of the solution is approximately equal to that of pure water, which is 23 mm Hg in this case.

Thus, the vapour pressure of the solution is close to 23 mm Hg.