What is the vapor pressure lowering , D P, for an aqueous solution prepared by mixing 50.0 grams of glycerol (non-volatile solute) and 500 grams of H₂O at 25 ° C?
Molar Mass(glycerol) = 92.09 g/mol and Molar Mass(H₂O) = 18.02 g/mol; P_H2O(25 ° C) = 23.76 torr.

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Quizplus · Accepted Answer

The vapor pressure lowering, ΔP, can be calculated using 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 glycerol and water:- Moles of glycerol = $ \frac{50.0 \, \text{g}}{92.09 \, \text{g/mol}} \approx 0.543 \, \text{mol} $- Moles of water = $ \frac{500 \, \text{g}}{18.02 \, \text{g/mol}} \approx 27.75 \, \text{mol} $The mole fraction of water, $ X_{\text{water}} $, is calculated as:$$ X_{\text{water}} = \frac{\text{moles of water}}{\text{moles of water} + \text{moles of glycerol}} = \frac{27.75}{27.75 + 0.543} \approx 0.981 $$The vapor pressure lowering, $ \Delta P $, is then:$$ \Delta P = P^0_{\text{water}} - (X_{\text{water}} \times P^0_{\text{water}}) $$$$ \Delta P = 23.76 \, \text{torr} - (0.981 \times 23.76 \, \text{torr}) \approx 23.76 \, \text{torr} - 23.3 \, \text{torr} = 0.456 \, \text{torr} $$Therefore, the correct answer is $ \Delta P = 0.456 \, \text{torr} $, which corresponds to choice A.

What Is the Vapor Pressure Lowering , D P, for an Aqueous