The solvent for an organic reaction is prepared by mixing 50.0 mL of acetone ( C
ID: 483331 • Letter: T
Question
The solvent for an organic reaction is prepared by mixing 50.0 mL of acetone ( C 3 H 6 O ) with 48.0 mL of ethyl acetate ( C 4 H 8 O 2 ). 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?
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
= 50.0 x 0.791 = 39.55 g
Moles of acetone = mass/molar mass
= 39.55/58.08 = 0.6809 mol
Mass of ethyl acetate = volume x density
= 48.0 x 0.900 = 43.2 g
Moles of ethyl acetate = mass x molar mass
= 43.2/88.105 = 0.4903 mol
Mole fraction of acetone x(acetone) = 0.6809/(0.6809 + 0.4903) = 0.5813
Mole fraction of ethyl acetate x(ethyl acetate) = 0.4903/(0.6809 + 0.4903) = 0.4186
From Raoult's law
Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)
= 0.5813 x 230.0 + 0.4186 x 95.38 = 173.62 mm Hg