Question

At 25°C the saturated vapour pressure of water is 24 mm Hg. Find the saturated vapour pressure of a 5% aqueous solution of urea at the same temperature. (Molar mass of urea = 60 g mol$^{-1}$)

Hint:

Raoult's Law is helpful for determining the effect of solutes on the vapour pressure of solutions. The greater the mole fraction of solvent, the closer the solution’s vapour pressure will be to that of the pure solvent.

Answer 1

According to Raoult’s Law, the vapour pressure of a solution is given by: \[ P_{\text{solution}} = P_{\text{solvent}} \times X_{\text{solvent}} \] Where \( P_{\text{solution}} \) is the vapour pressure of the solution, \( P_{\text{solvent}} \) is the vapour pressure of the pure solvent, and \( X_{\text{solvent}} \) is the mole fraction of the solvent. For a 5% (w/v) solution of urea: \[ X_{\text{solvent}} = \frac{\text{moles of solvent}}{\text{moles of solvent} + \text{moles of solute}} \] First, calculate the moles of urea (solute) and the moles of water (solvent). Given: - Mass of urea = 5 g - Molar mass of urea = 60 g/mol \[ \text{Moles of urea} = \frac{5}{60} = 0.0833 \, \text{mol} \] The mass of water in the solution is \( 100 \, \text{g} - 5 \, \text{g} = 95 \, \text{g} \). Molar mass of water = 18 g/mol \[ \text{Moles of water} = \frac{95}{18} = 5.28 \, \text{mol} \] Now, calculate the mole fraction of the solvent (water): \[ X_{\text{water}} = \frac{5.28}{5.28 + 0.0833} = 0.984 \] Thus, the vapour pressure of the solution is: \[ P_{\text{solution}} = 24 \, \text{mm Hg} \times 0.984 = 23.6 \, \text{mm Hg} \]