What is the vapor pressure of water above a solution prepared by adding 15.0 grams of lactose (C₁₂H₂₂O₁₁) to 100.0 grams of water at 338 K?
The vapor pressure of pure water at 338 K equals 187.5 torr. Molar Mass(C₁₂H₂₂O₁₁) = 342.30 g/mol and Molar Mass(H₂O) = 18.015 g/mol.

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Quizplus · Accepted Answer

To find the vapor pressure of water above the solution, we use Raoult's law, which states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution. First, calculate the moles of lactose and water:- Moles of lactose = $ \frac{15.0 \, \text{g}}{342.30 \, \text{g/mol}} $- Moles of water = $ \frac{100.0 \, \text{g}}{18.015 \, \text{g/mol}} $Then, calculate the mole fraction of water and apply Raoult's law:- Mole fraction of water = $ \frac{\text{moles of water}}{\text{moles of water} + \text{moles of lactose}} $- Vapor pressure of the solution = $ \text{Vapor pressure of pure water} \times \text{Mole fraction of water} $Given that the vapor pressure of pure water at 338 K is 187.5 torr, and considering the small amount of lactose compared to water, the vapor pressure of the solution will be slightly less than that of pure water but still very close to 187.5 torr, making 186 torr the most reasonable answer.

What Is the Vapor Pressure of Water Above a Solution