The solvent for an organic reaction is prepared by mixing 70.0 m L of acetone (

Question

The solvent for an organic reaction is prepared by mixing 70.0mL of acetone (C3H6O) with 61.0mL 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. I got 180.88 but they're saying it is not right. The solvent for an organic reaction is prepared by mixing 70.0mL of acetone (C3H6O) with 61.0mL 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. I got 180.88 but they're saying it is not right. Compound Vapor pressure
(mmHg) Density
(g/mL) acetone 230.0 0.791 ethyl acetate 95.38 0.900

Explanation / Answer

Basically, the same thing is done here; except we find the mole fractions for both compounds and then add them together.
So first find the moles; we can do this through the density given.
acetone= (0.791 g/mL x 70 mL)
= 55.37 g
moles = 55.37 g / 58.079 g/mol
= 0.953 moles

ethyl acetate = (0.900g/mL x 61 mL)
= 54.9 g
moles = 54.9 g / 88.105 g/mol
= 0.623 moles

Now, we figure out the mole fractions:
mole fraction of acetone = (0.953 / (0.953 + 0.623))
= 0.605
mole fraction of ethyl acetate = (0.623 / (0.953 + 0.623))
= 0.395

Now; we solve for Psoln
Psoln= (0.605 x 230) + (0.395 x 95.38)
= 176.8 mmHg

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