High School

What is the vapor pressure of the solution prepared by mixing 3.0 mol of [tex]C_2H_5OH[/tex] and 2.0 mol of [tex]C_3H_7OH[/tex]? Assume ideal behavior.

Given:
- [tex]P^\circ_{C_2H_5OH} = 100[/tex] torr
- [tex]P^\circ_{C_3H_7OH} = 40[/tex] torr

Select one:
a. 100.2 torr
b. 76 torr
c. 420 torr
d. 140 torr

Answer :

Final answer:

The vapor pressure of the solution prepared by mixing 3.0 mol C2H5OH and 2.0 mol C3H7OH is 76 torr.

Explanation:

To calculate the vapor pressure of the solution, we can use Raoult's law. Raoult's law states that the vapor pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapor pressure in its pure state.

First, let's calculate the mole fraction of each component:

Mole fraction of C2H5OH = moles of C2H5OH / total moles of both components

Mole fraction of C2H5OH = 3.0 mol / (3.0 mol + 2.0 mol) = 0.6

Mole fraction of C3H7OH = moles of C3H7OH / total moles of both components

Mole fraction of C3H7OH = 2.0 mol / (3.0 mol + 2.0 mol) = 0.4

Next, we can calculate the vapor pressure of the solution:

Vapor pressure of the solution = (mole fraction of C2H5OH) * (vapor pressure of C2H5OH) + (mole fraction of C3H7OH) * (vapor pressure of C3H7OH)

Vapor pressure of the solution = (0.6) * (100 torr) + (0.4) * (40 torr)

Vapor pressure of the solution = 60 torr + 16 torr = 76 torr

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