What is the vapor pressure of water above an aqueous solution prepared by mixing 300 grams of glycerol (non-volatile solute) and 505 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

Question

Quizplus · Accepted Answer

Using Raoult's law, we can calculate the vapor pressure of the solution as follows: P(H2O) = X(H2O) * P(H2O)^° where X(H2O) is the mole fraction of water in the solution and P(H2O)^° is the vapor pressure of pure water at the given temperature. Since glycerol and H₂O are both non-volatile solutes, their mole fractions can be assumed to be negligible. The mole fraction of water can be calculated as follows: moles(H2O) = 505 g / 18.02 g/mol = 28.0 mol moles(glycerol) = 300 g / 92.09 g/mol = 3.26 mol moles(total) = moles(H2O) + moles(glycerol) = 31.3 mol X(H2O) = moles(H2O) / moles(total) = 0.893 Using the given value for P(H2O)^°, we get: P(H2O) = 0.893 * 23.76 torr = 21.3 torr Therefore, the vapor pressure of water above the solution is 21.3 torr. Choice C is the correct answer.

What Is the Vapor Pressure of Water Above an Aqueous