Part A Part B The solvent for an organic reaction is prepared by mixing 70.0 mL
ID: 485999 • Letter: P
Question
Part A
Part B
The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 71.0 mLof ethyl acetate (C4H8O2). This mixture is stored at 25.0 C. The vapor pressure and the densities for the two pure components at 25.0 C are given in the following table. What is the vapor pressure of the stored mixture?
Express your answer to three significant figures and include the appropriate units.
Compound Vapor pressure(mmHg) Density
(g/mL) acetone 230.0 0.791 ethyl acetate 95.38 0.900
Explanation / Answer
Mass of acetone = volume x density
= 70.0 x 0.791 = 55.37 g
Moles of acetone = mass/molar mass
= 55.37/58.08 = 0.9533 mol
Mass of ethyl acetate = volume x density
= 71.0 x 0.900 = 69.58 g
Moles of ethyl acetate = mass x molar mass
= 69.58/88.105 = 0.7897 mol
Mole fraction of acetone x(acetone) = 0.9533/(0.9533 + 0.7897) = 0.5469
Mole fraction of ethyl acetate x(ethyl acetate) = 0.7897/(0.9533 + 0.7897) = 0.4530
From Raoult's law
Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)
= 0.5469 x 230.0 + 0.4530 x 95.38 = 168.99 mm Hg
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