Question

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

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.

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

Answer with three sig figs: 176 mmHg