Question

The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 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?

Answer 1

Answer:

The answer is "170.9 mm Hg".

Explanation:

$\text{Mass of acetone = volume} \times density$

                          $= 70.0 \times 0.791\\\\ = 55.37\ g$

$\text{Moles of acetone} = \frac{mass}{molar\ mass}$

                            $=\frac{55.37}{58.08}\\\\ = 0.9533\ mol$

$\text{Mass of ethyl acetate = volume} \times density$

                                   $= 73.0 \times 0.900\\\\ = 65.7\ g$  

$\text{Moles of ethyl acetate = mass} \times\ molar\ mass$

                                    $= \frac{65.7}{88.105} \\\\= 0.7457\ mol$

$\text{Mole fraction of acetone x(acetone)} = \frac{0.9533}{(0.9533 + 0.7457)}\ = 0.5611$ $\text{Mole fraction of ethyl acetate x(ethyl acetate)} =\frac{0.7457}{(0.9533 + 0.7457) }= 0.4389$

Applying Raoult's law: $\text{Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)}\\\\= 0.5611 \times 230.0 + 0.4389 \times 95.38\\\\ = 170.9\ mm \ Hg$

Answer 2

The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2).

The vapor pressure of the stored mixture is: 170.03 mmHg

In the given information, there is some information that is still missing.

The parameters that we are being given include:

• The volume of acetone = 70.0 mL
• The volume of ethyl acetate = 75.0 mL
• The standard temperature for the mixture = 25° C

The  first step we need to take is to determine the mass and number of moles  of each compound (i.e. for acetone and ethyl acetate)

For us to do that:

We need the density of acetone and ethyl acetate, which is not given:

Assuming that at a standard condition of vapour pressure:

• 230 mmHg of acetone has a density of 0.791 g/mL
• 95.38 mmHg of ethyl acetate has a density of 0.900 g/mL

Then;

Using the relation:

$\mathbf{Density = \dfrac{Mass}{volume}}$

Mass of acetone = Density of acetone × volume of acetone

Mass of acetone = 0.791  g/mL  × 70.0 mL

Mass of acetone = 55.37 g

Mass of ethyl acetate = Density of ethyl acetate  × volume of ethyl acetate

Mass of ethyl acetate = 0.900 g/mL  ×  75.0 mL

Mass of ethyl acetate = 67.5 g

At standard conditions;

• For acetone, molar mass = 58.08 g/mol
• For ethyl acetate, molar mass = 88.11 g/mol

Now, using the formula for calculating the numbers of moles which can be expressed as:

$\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}$

For acetone:

$\mathbf{Number \ of \ moles = \dfrac{55.37 \ g}{58.08 \ g/mol}}$

$\mathbf{Number \ of \ moles =0.95334 \ mol}$

For ethyl acetate:

$\mathbf{Number \ of \ moles = \dfrac{67.5 \ g}{88.11 \ g/mol}}$

$\mathbf{Number \ of \ moles =0.76609 \ mol}$

Now, we will determine the mole fraction of each compound.

The mole fraction describes the ratio a certain constituent of a mixture to the total amount of all the constitutent in the mixture.

Using the formula:

$\mathbf{mole \ fraction = \dfrac{n_A}{n_A+n_B+...n_N}}$

For Acetone:

$\mathbf{mole \ fraction = \dfrac{0.95334}{0.95334+0.76609}}$

$\mathbf{mole \ fraction =0.5545 }$

For ethyl acetate:

$\mathbf{mole \ fraction = \dfrac{0.76609}{0.76609+0.95334}}$

$\mathbf{mole \ fraction =0.4455}$

Finally, we can compute determine the vapour pressure of the stored mixture using Raoult's Law.

Raoult's Law posits that the constituent of a partial pressure in a mixture of a liquid is proportional to the mole fraction of that constituent in the mixture provided the temperature is constant.

∴ For the stored mixture = Vapor pressure of acetone + vapour pressure of ethyl acetate.

where:

• Vapour pressure of the solution = (mole fraction × vapor pressure) of solvent

For acetone;

Vapor pressure = 0.5545 × 230 mmHg

Vapour pressure = 127.54 mmHg

For ethyl acetate:

Vapour pressure = 0.4455 × 95.38 mmHg

Vapour pressure ==42.49 mmHg

Thus, the vapor pressure of the stored mixture is

= (127.54 + 42.49 ) mmHg

= 170.03 mmHg

Therefore, we can conclude that the vapour pressure of the stored mixture is 170.03 mmHg

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