Question

A solution containing 2 g of glucose (M = 180 g mol$^{-1}$) in 100 g of water is prepared at 303 K. If the vapour pressure of pure water at 303 K is 32.8 mm Hg, what would be the vapour pressure of the solution?

Hint:

Raoult's Law states that the vapour pressure of a solvent in a solution is proportional to the mole fraction of the solvent. The presence of a solute lowers the vapour pressure.

Answer 1

We can calculate the vapour pressure of the solution using Raoult’s Law, which states that the vapour pressure of the solvent in the solution is proportional to the mole fraction of the solvent. \[ P = P_0 \times X_{\text{solvent}} \] Where: - \( P_0 = 32.8 \, \text{mm Hg} \) is the vapour pressure of pure water,
- \( X_{\text{solvent}} \) is the mole fraction of water.
Step 1: Calculate the moles of glucose: \[ \text{Moles of glucose} = \frac{2}{180} = 0.0111 \, \text{mol} \] Step 2: Calculate the moles of water: \[ \text{Moles of water} = \frac{100}{18} = 5.56 \, \text{mol} \] Step 3: Calculate the mole fraction of water: \[ X_{\text{water}} = \frac{\text{moles of water}}{\text{moles of water} + \text{moles of glucose}} = \frac{5.56}{5.56 + 0.0111} = 0.998 \] Step 4: Calculate the vapour pressure of the solution: \[ P = 32.8 \times 0.998 = 32.734 \, \text{mm Hg} \] Thus, the vapour pressure of the solution is 32.734 mm Hg.