A sample of 20.0 g of a nonvolatile, nonelectrolyte (molar mass = 120 g/mol) is dissolved in 10.0 grams of ethanol (C₂H₅OH). If the vapor pressure of pure ethanol at the temperature of the experiment is 0.250 atm, what is the vapor pressure of ethanol above the solution?

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Quizplus · Accepted Answer

The vapor pressure of the solution can be calculated using Raoult's Law, which states that the vapor pressure of a solvent above a solution (P_solution) is equal to the vapor pressure of the pure solvent (P_pure) multiplied by the mole fraction of the solvent in the solution (X_solvent). First, calculate the moles of solute (nonvolatile, nonelectrolyte) and solvent (ethanol).Moles of solute = mass / molar mass = 20.0 g / 120 g/mol = 0.167 molMoles of ethanol = mass / molar mass = 10.0 g / 46.07 g/mol (molar mass of ethanol) = 0.217 molTotal moles = moles of solute + moles of ethanol = 0.167 mol + 0.217 mol = 0.384 molMole fraction of ethanol = moles of ethanol / total moles = 0.217 mol / 0.384 mol = 0.565Vapor pressure of ethanol above the solution = P_pure * X_solvent = 0.250 atm * 0.565 = 0.141 atm

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