Question

The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 56.0 mL 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.

Answer 1

First, we will convert the two substances' volumes into grams using the provided densities.

Acetone: 70.0 mL x (0.791 g/mL) = 55.37 grams of Acetone

Ethyl Acetate: 56.0 mL x (0.900 g/mL) = 50.4 grams of Ethyl Acetate

Convert these quantities to moles by dividing the mass by the molar mass:

Acetone: 55.37 g / (58.08034 g/mol) = 0.9533 moles Acetone

Ethyl Acetate: 50.4 g / (88.1063 g/mol) = 0.572 moles Ethyl Acetate

Now we determine the mole fraction for each substance:

Acetone: 0.9533 moles / (0.9533 moles + 0.572 moles) = 0.625

Ethyl Acetate: 0.572 moles / (0.9533 moles + 0.572 moles) = 0.375

Finally, we apply Raoult's Law: multiply the mole fraction by the vapor pressure from the table:

Acetone: 0.625 x 230.0 mmHg = 143.75 mmHg

Ethyle Acetate: 0.375 x 95.38 mmHg = 35.7675 mmHg

Add the two vapor pressures:

Answer: 179.5175 mmHg

Answer with three sig figs: 176 mmHg