Question

Safrole was once used as a flavoring in root beer, until it was banned in 1960. what is the vapor pressure of a solution prepared by dissolving 0.75 mol of nonvolatile safrole in 950 g of ethanol (46.07 g/mol)? p°ethanol = 50.0 torr at 25°c.

Answer 1

Vapor pressure of solution = 48.3 torr

Step-by-step explanation:

Given:

Moles of safrole (solute) = 0.75

Mass of ethanol (solvent) = 950 g

Vapor pressure of ethanol = 50.0 torr

To determine:

Vapor pressure of the solution

Step-by-step explanation:

Based on Raoult's Law, vapor pressure of a solution is expressed as:

`P = X(solvent) * P^(0) (solvent)`

`where\ X = mole\ fraction\ of\ solvent\\\\X = (moles\ of\ solvent)/(moles(solvent+solute)) \\\\P^(0) = vapor pressure of pure solvent`

moles of
`moles\ ethanol\ solvent = (950)/(46.07) =20.6moles\\\\moles\ safrole\ solute = 0.75 mol\\\\X(ethanol) = (20.6)/(21.35) =0.965\\\\P(solution) = 0.965*50.0 = 48.3 torr`

Answer 2

Answer is: 48,25 torr.
Raoult's Law: p = x(solv) · p(solv)
p - vapour pressure of a solution.
x(solv) - mole fraction of the solvent.
p(solv) - vapour pressure of the pure solvent.
n(ethanol) = 950g ÷ 46,07g/mol = 20,62 mol.
x(solv) = moles of solvent ÷ total number of moles
x(solv) = 20,62 ÷ 21,77 = 0,965.
p = 0,965 ·50,0 torr = 48,25 torr.