What is the vapor pressure in mm Hg of a solution of 16.0 g of glucose (C6H12O6) in 80.0 g of methanol (CH3OH) at 27 °C? The vapor pressure of pure methanol at 27 °C is 140 mm Hg.

Determine the mole fraction of methanol in the solution. The mole fraction of methanol (\( X_{\text{methanol}} \)) is given by: \( X_{\text{methanol}} = \frac{\text{moles of methanol}}{\text{moles of methanol} + \text{moles of glucose}} \).

Apply Raoult's Law to find the vapor pressure of the solution. Raoult's Law states that the vapor pressure of the solution (\( P_{\text{solution}} \)) is equal to the mole fraction of the solvent (methanol) times the vapor pressure of the pure solvent: \( P_{\text{solution}} = X_{\text{methanol}} \times P^0_{\text{methanol}} \), where \( P^0_{\text{methanol}} \) is the vapor pressure of pure methanol.