Question

At 20 °C, the vapour pressure of pure ethanol is 32.1 torr and the vapour pressure of pure methanol is 86.2 torr. Assuming ideal behaviour what is the vapour pressure (in torr) at 20 °C of a solution prepared by mixing 2.5 moles of methanol and 3.3 moles of ethanol?

Answer 1

n(ethanol),n1 = 3.3 mol
n(methanol),n2 = 2.5 mol

Total number of mol = n1+n2
= 3.3 + 2.5
= 5.8 mol

use:
Mole fraction of each components are
X(ethanol) = n1/total mol
= 3.3/5.8
= 0.569

use:
X(methanol) = n2/total mol
= 2.5/5.8
= 0.431


According to Raoult’s law:
P(ethanol) = Po(ethanol)*X(ethanol)
P(ethanol) = 32.1 torr*0.569
P(ethanol) = 18.2638 torr


According to Raoult’s law:
P(methanol) = Po(methanol)*X(methanol)
P(methanol) = 86.2 torr*0.431
P(methanol) = 37.1552 torr

total pressure = sum of individual pressures
total pressure = 18.2638 torr + 37.1552 torr
total pressure = 55.419 torr
Answer: 55.4 torr