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

Answer:

Vapor pressure of solution = 48.3 torr

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

Explanation:

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

$P = X(solvent) * P^{0} (solvent)$

$where\ X = mole\ fraction\ of\ solvent\\\\X = \frac{moles\ of\ solvent}{moles(solvent+solute)} \\\\P^{0} = vapor pressure of pure solvent$

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

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.