Question

At 365 K, pure toluene and hexane have vapor pressures of 5.82*10⁴ Pa and 1.99*10⁵ Pa, respectively.

a. Calculate the mole fraction of hexane in the liquid mixture that boils at 365 K at a pressure of 1140 Torr.

b. Calculate the mole fraction of hexane in the vapor that is in equilibrium with the liquid of part (a)

Answer 1

According to rault's law : P(toluene) = x1 * P₁^0 and P(hexane) = x2 * P₂^0

where P1^0 and P2^0 is the vapour pressure of pure toluene and hexane respectively. x1 and x2 is the mole fraction of toluene and hexane in the mixture.

Total pressure of the mixture = x1 * P₁^0 + x2 * P₂^0

A lliquid boils when the total pressure becomes equal to the 1140 torr

1140 torr = 151987.5Pa

Since molefraction adds up to one , it can be expressed as : x1 = 1-x2. Thus the above equation can be written as follows.

Total pressure of the mixture = x1 * P₁^0 + x2 * P₂^0

or, 151987.5 Pa   = (1-x2)*P₁^0 + x2 * P₂^0

or, 151987.5 Pa= (1-x2)*5.82*10^4 + x2*1.99*10^5

or, x2 =151987.5 Pa-1.99*10^5/5.82*10^4-1.99*10^5 = 0.33

Mole fraction of hexane = 0.33

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The mole fraction of a component in an ideal gas mixture equals its partial pressure divided by the total pressure. The partial pressure in vapor-liquid-equlibrium state is nothing else but the vapor pressure of the component.
So the mole fraction of hexane in vapor phase is:

x = 0.33 *1.99*10^5/151987.5 = 0.44