What is the vapor pressure of a 45.0 % solution of glucose, C₆H₁₂O₆, at 90.0^$\circ$C, given that the vapor pressure of pure water at that temperature is 526 mm Hg?

Question

Quizplus · Accepted Answer

To find the vapor pressure of the solution, we can use Raoult's Law, which states that the vapor pressure of a solution is equal to the mole fraction of the solvent times the vapor pressure of the pure solvent. Since glucose is a non-volatile solute, it will not contribute to the vapor pressure directly. The mole fraction of water (solvent) can be calculated by considering the percentage of the solution that is water, which is 100% - 45% = 55%. Assuming ideal behavior and that the percentage is a rough approximation of the mole fraction (which is a simplification because the question does not provide enough information to calculate exact mole fractions), the vapor pressure of the solution is approximately 55% of the vapor pressure of pure water. Therefore, 0.55 * 526 mm Hg = 289.3 mm Hg, which is closest to option A, 486 mm Hg, given the options provided. However, it seems there might be a misunderstanding in the calculation or the interpretation of the options. Based on the calculation, none of the options accurately match, indicating a potential error in the explanation or a misinterpretation of the question's requirements. The correct approach involves using Raoult's Law and the given data, but the precise answer seems to be mismatched with the provided options.

What Is the Vapor Pressure of a 45 ∘\circ∘ C, Given That the Vapor Pressure of Pure Water at of Glucose

What Is the Vapor Pressure of a 45 $\circ$ C, Given That the Vapor Pressure of Pure Water at of Glucose