Question

"a solution is made by mixing 44.0 ml of ethanol, c2h6o, and 56.0 ml of water. assuming ideal behavior, what is the vapor pressure of the solution at 20 °c

Answer 1

Vapor pressure of the solution = Partial pressure of ethanol + partial pressure of water.

Partial pressure of ethanol can be calculated from the molefraction of ethanol and vapor pressure of pure ethanol at
`20^(0)C`

Density of ethanol = 0.800 g/mL

Finding out mass of ethanol =
`44.0 mL * (0.800 g)/(1 mL) = 35.2 g`

Moles of ethanol =
`35.2 g * (1 mol)/(46.07 g) =0.764 mol`

Density of water = 1.00 g/mL

Mass of water =
`56.0 mL * (1.00 g)/(mL) = 56.0 g`

Moles of water =
`56.0 g * (1 mol)/(18.02 g) = 3.11 mol`

Total number of moles of the solution = 0.764 mol + 3.11 mol = 3.874 mol

Molefraction of ethanol =
`(0.764 mol)/(3.874 mol) = 0.197`

Molefraction of water =
`(3.11 mol)/(3.874 mol) = 0.803`

Vapor pressure of pure ethanol at
`20^(0)C = 59.3 mm Hg`

Vapor pressure of pure water at
`20^(0)C = 17.5 mmHg`

Using Raoult's law to find out partial pressures:

`P_(ethanol) = (Molefraction_(ethanol)) P^(0)_(ethanol)`

`P_(ethanol)= 0.197 (59.3 mmHg)` = 11.68 mmHg

`P_(water) = (Molefraction)_(water)(P^(0)_(water))`

`P_(water) =`
`0.803 (17.5 mmHg)` = 14.05 mmHg

Total vapor pressure of the solution = 11.68 mmHg + 14.05 mmHg

= 25.73 mmHg