Question

At 20 ∘C the vapor pressure of benzene C6H6 is 75 torr and that of toluene C7H8 is 22 torr. Assume that benzene and toluene form an ideal solution.

a) In a solution composed of benzene and toluene that has a vapor pressure of 37 torr at 20 ∘C, what is the mole fraction of benzene?

b) In a solution composed of benzene and toluene that has a vapor pressure of 37 torr at 20 ∘C, what is the mole fraction of toluene?

Answer 1

a) Xbenzene = 0.283

b) Xtoluene = 0.717

Step-by-step explanation:

At T = 20°C:

⇒ vapor pressure of benzene (P*b) = 75 torr

⇒ vapor pressure toluene (P*t) = 22 torr

Raoult's law:

• Pi = Xi.P*i

∴ Pi: partial pressure of i

∴ Xi: mole fraction

∴ P*i: vapor pressure at T

a) solution: benzene (b) + toluene (t)

∴ Psln = 37 torr; at T=20°C

⇒ Psln = Pb + Pt

∴ Pb = (Xb)*(P*b)

∴ Pt = (Xt)*(P*t)

∴ Xb + Xt = 1

⇒ Psln = 37 torr = (Xb)(75 torr) + (1 - Xb)(22 torr)

⇒ 37 torr - 22 torr = (75 torr)Xb - (22 torr)Xb

⇒ 15 torr = 53 torrXb

⇒ Xb = 15 torr / 53 torr

⇒ Xb = 0.283

b) Xb + Xt = 1

⇒ Xt = 1 - Xb

⇒ Xt = 1 - 0.283

⇒ Xt = 0.717