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

The vapour pressure of a solution can be determined using Raoult’s Law. This law posits that the vapour pressure of the solvent within a solution is directly proportional to its mole fraction. The formula is given by: \[ P = P_0 \times X_{\text{solvent}} \] Where: - \( P_0 = 32.8 \, \text{mm Hg} \) represents the vapour pressure of pure water. - \( X_{\text{solvent}} \) denotes the mole fraction of water in the solution. To calculate the vapour pressure of the solution, follow these steps: Step 1: Determine 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: Compute 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. \[ P = 32.8 \times 0.998 = 32.734 \, \text{mm Hg} \] Consequently, the vapour pressure of the solution is 32.734 mm Hg.