The solvent for an organic reaction is prepared by mixing 50.0 of acetone () with 57.0 of ethyl acetate (). This mixture is stored at 25.0 . The vapor pressure and the densities for the two pure components at 25.0 are given in the following table. What is the vapor pressure of the stored mixture?

Question

The solvent for an organic reaction is prepared by mixing 50.0 of acetone () with 57.0 of ethyl acetate (). This mixture is stored at 25.0 . The vapor pressure and the densities for the two pure components at 25.0 are given in the following table. What is the vapor pressure of the stored mixture?

Answer 1

To determine the vapor pressure of the stored mixture, we need to calculate the composition of the mixture and then use Raoult's law to find the vapor pressure.

1. Calculate the mole fractions of the components:
The mole fraction (X) of a component can be calculated by dividing the number of moles of that component by the total number of moles in the mixture.

Number of moles of acetone (n1) = mass of acetone (m1) / molar mass of acetone (M1)
Number of moles of ethyl acetate (n2) = mass of ethyl acetate (m2) / molar mass of ethyl acetate (M2)

Given:
mass of acetone (m1) = 50.0 g
mass of ethyl acetate (m2) = 57.0 g
molar mass of acetone (M1) = 58.08 g/mol
molar mass of ethyl acetate (M2) = 88.11 g/mol

n1 = 50.0 g / 58.08 g/mol = 0.861 mol
n2 = 57.0 g / 88.11 g/mol = 0.647 mol

Total number of moles (n_total) = n1 + n2 = 0.861 mol + 0.647 mol = 1.508 mol

The mole fractions of acetone and ethyl acetate are:
X1 (acetone) = n1 / n_total = 0.861 mol / 1.508 mol ≈ 0.571
X2 (ethyl acetate) = n2 / n_total = 0.647 mol / 1.508 mol ≈ 0.429

2. Use Raoult's law to find the vapor pressure:
Raoult's law states that the vapor pressure of a component in a mixture is equal to the product of its mole fraction and the vapor pressure of the pure component.

Vapor pressure of the mixture (P_mixture) = X1 * P1 + X2 * P2

Given:
Vapor pressure of acetone (P1) = 253 torr
Vapor pressure of ethyl acetate (P2) = 77 torr (rounded from the table)

P_mixture = 0.571 * 253 torr + 0.429 * 77 torr
P_mixture ≈ 229.223 + 33.233
P_mixture ≈ 262.456 torr

Therefore, the vapor pressure of the stored mixture is approximately 262.456 torr.

Answer 2

You must have units. 50.0 what? 57.0 what? 25.0 what?

Look up the density in the table to which you refer. Use the density to convert mL to grams. Then moles = grams/molar mass.
mole fraction X = moles X/total moles.

Finally,
Pacetone = Xacetone*P^o_acetone and
Petacetate = Xetacetate*P^{o_{ethyl acetate}

Then add Pacetone + Pethyl aectate.}