The solvent for an organic reaction is prepared by mixing 50.0 mL of acetone (C3
ID: 984863 • Letter: T
Question
The solvent for an organic reaction is prepared by mixing 50.0 mL of acetone (C3H6O) with 49.0 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
Compound
Vapor pressure
(mmHg)
Density
(g/mL)
acetone
230.0
0.791
ethyl acetate
95.38
0.900
Explanation / Answer
Density of Acetone = Mass of Acetone/Volume of acetone
Mass of Acetone = 0.791 gm/ml * 50 ml = 39.55 gms
Molar mass of Acetone(C3H6O) = 3 * 12 + 6 * 1 + 16 = 36 + 22 = 58 gm/mol
Density of Ethyl Acetate = Mass of Ethyl Acetate/Volume of Ethyl Acetate
Mass of Ethyl Acetate = 0.900 gm/ml * 49 ml = 44.1 gms
Molar mass of Ethyl Acetate = 4 * 12 + 8 * 1 + 2 * 16 = 88 gm/mol
Number of moles of acetone = 39.55/58 = 0.6818 moles
Number of moles of Ethyl Acetate = 44.1/88 = 0.5011 moles
Mole fraction of acetone = 0.6818/(0.6818+0.5011) = 0.5736
Mole fraction of Ethyl acetate = 1 - 0.5736 = 0.4236
Pressure of Solution = 0.4236 * 95.38 + 0.5736 * 230
=> 172.55 mm Hg