The solvent for an organic reaction is prepared by mixing 70.0 m L of acetone (
ID: 823769 • Letter: T
Question
The solvent for an organic reaction is prepared by mixing 70.0mL of acetone (C3H6O) with 75.0mL of 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? Compound Vapor pressure(mmHg) Density
(g/mL) acetone 230.0 0.791 ethyl acetate 95.38 0.900 Express your answer to three significant figures and include the appropriate units. The solvent for an organic reaction is prepared by mixing 70.0mL of acetone (C3H6O) with 75.0mL of 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? Compound Vapor pressure
(mmHg) Density
(g/mL) acetone 230.0 0.791 ethyl acetate 95.38 0.900 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
Moles of acetone = mass/molar mass = volume x density/molar mass of acetone
= 70.0 x 0.791/58.08 = 0.95334 mol
Moles of ethyl acetate = mass/molar mass = volume x density/molar mass of ethyl acetate
= 75.0 x 0.900/88.105 = 0.76613 mol
P(solution) = mole fraction of acetone x vapor pressure of acetone + mole fraction of ethyl acetate x vapor pressure of ethyl acetate
= 0.95334/(0.95334 + 0.76613) x 230.0 + 0.76613/(0.95334 + 0.76613) x 95.38
= 170 mmHg