A liquid mixture at equilibrium with its vapor at 25 ºC comprises 35.8 g of benz

Question

A liquid mixture at equilibrium with its vapor at 25 ºC comprises 35.8 g of benzene (subscript A) and 56.7 g of toluene (subscript B) in the liquid phase. At that temperature the vapor pressure of liquid benzene pure is PAº = 95.1 Torr and the vapor pressure of liquid toluene pure is PBº = 28.4 Torr. The corresponding molar masses of the compounds are MA = 78.11 g / mol and MB = 92.13 g / mol. Assume that benzene and toluene form ideal solutions upon mixing.

(a) What is the vapor pressure of the mixture?

(b) What is the mol fraction of benzene in the vapor phase?

Explanation / Answer

(a) Moles of benzene in the mixture, mA = (35.8 gm)*(1 mole/78.11 gm) = 0.4583

Moles of toluene in the mixture, mB = (56.7 gm)*(1 mole/92.13 gm) = 0.6154

Mole fraction of benzene in the mixture, xA = mA/(mA + mB) = 0.4583/(0.4583 + 0.6154) = 0.4268

Mole fraction of toluene in the mixture, xB = mB/(mA + mB) = 0.6154/(0.4583 + 0.6154) = 0.5731

As per Raoult’s law,

PA = xAPA0 = (0.4268)*(95.1 torr) (the vapour pressure of pure benzene is given as 95.1 torr).

Or, PA = 40.58868 torr

Similarly, PB = xBPB0 = (0.5731)*(28.4 torr) = 16.27604 torr

The total vapour pressure of the mixture is P = PA + PB = (40.58868 + 16.27604) torr = 56.86472 torr 56.86 torr (ans)

(b) The mole fraction of benzene in the vapour phase is given by

yA = xAPA0/[PB0 + (PA0 – PB0)xA]

=è yA = [(0.4268)*(95.1)]/[(28.4) + (95.1 – 28.4)*(0.4268]

=è yA = (40.58868)/[28.4 + (95.1 – 28.4)*(0.4268)]

=è yA = 40.58868/56.86756 = 0.7137 0.71 (ans)

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