Question

A solution of benzene and toluene at 25 °C exerts a total vapor pressure of 66.6 torr. What is the mole fraction of benzene in the vapor phase? The vapor pressures of pure benzene and toluene at 25 °C are 94.2 torr and 28.4 torr, respectively. 0

Answer 1

Answer:

0.5805.

Explanation:

Using Raoult's Law;

Psol = Xbenzene * P°benzene + Xtoluene * P°toluene

where,

Psol = vapor pressure of the solution

Xbenzene = mole fraction of benzene

P°benzene = vapor pressure of pure benzene

Xtoluene = mole fraction of toluene

P°toluene = vapor pressure of pure toluene

Mole fraction is defined as the ratio between the number of moles of a component in a solution to the total number of moles in the solution.

Xbenzene + Xtoluene = 1

Let,

Y = mole fraction of toluene

X = mole fraction of benzene

X = 1 − Y

Psol = X * P°benzene + (1 − X) * P°toluene

66.6 = X * 94.2 + (1 − X) * 28.4

X = 0.5805

Therefore,

Y = 1 - 0.5805

= 0.4195

Mole fraction of benzene = 0.5805.

Answer 2

The mole fraction of benzene in the vapor pressure is 0.5805 torr.

The pressure of the mixture has been the constituent of the vapor pressure of the individual gases in the solution.

The total pressure (P) of the mixture has been given as:

$P=P_A\;X_A\;+\;P_B\;X_B$

Computation for the mole fraction of Benzene

The total pressure of the mixture is $P=66.6\;\rm torr$

The vapor pressure of Benzene is $P_A=94.2\;\rm torr$

The vapor pressure of Toluene is $P_B=28.4\;\rm torr$

The total mole fraction of the benzene and toluene is 1.

Let, the mole fraction of benzene is $X_A=x$

The mole fraction of toluene is $X_B=1-x$

Substituting the values for the mole fraction:

$66.6=94.2\;\times x\;+\;28.4\;\times\;(1-x)\\66.6=94.2x\;-\;28.4x\;+\;28.4\\x=0.5805\;\rm torr$

Thus, the mole fraction of benzene in the vapor pressure is 0.5805 torr.

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