What is the vapor pressure (in mm Hg) at 30∘C of a solution prepared by dissolving 30.5 g of acetone in 23.5 g of ethyl acetate?

The vapor pressure of the solution is 284.29 mmHg

Explanation:

From Raoult's law

Vapor pressure of solution = mole fraction of solvent × vapor pressure of solvent

Mass of solute (acetone) = 30.5 g

MW of acetone (CH3COCH3) = 58 g/mol

Number of moles of solute = mass/MW = 30.5/58 = 0.526 mol

Mass of solvent (ethyl acetate) = 23.5g

MW of ethyl acetate (CH3COOC2H5) = 88 g/mol

Number of moles of solvent = mass/MW = 23.5/88 = 0.267 mol

Volume of solvent = 0.267×22.4×1000 = 5980.8 cm^3

Vapor pressure of solvent (P) = nRT/V

n is number of moles of solvent = 0.267 mol

R is gas constant = 82.057 cm^3.atm/mol.K

T is temperature of solution = 30°C = 30+273 = 303 K

V is volume of the solvent = 5980.8 cm^3

P = 0.267×82.057×303/5980.8 = 1.11 atm = 1.11×760 = 843.6 mmHg

Total moles of solution = moles of solute + moles of solvent = 0.526 + 0.267 = 0.793 mol

Mole fraction of solvent = moles of solvent/moles of solution = 0.267/0.793 = 0.337

Vapor pressure of solution = 0.337 × 843.6 mmHg = 284.29 mmHg

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