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The mole fraction of A (XA ) in the vapor phase of a mixture of two liquids is 0.24 and the sum of the partial pressures is 740 torr. Calculate the mol fraction of B (report to 2 decimal places) in liquid phase if the vapor pressure of pure B (PB0) 800 Torr. Show calculations.

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Answer:

Mole fraction of B is 0.76

Explanation:

The pressure of a mixture of gases using vapor pressure and mole fraction of each gas is:

P = P°aₓXa + P°bₓXb + ... + P°nₓXn

Where P is pressure of the system, P° is vapor pressure of pure gas and X is its mole fraction.

In the problem, the pressure of the mixture of two gases is 740torr, mol fraction of A is 0.24 and vapor pressure of B is 800torr, that is:

740torr =  P°aₓ0.24 + 800torrₓXb

Also, the sum of mole fractions for each of the compounds in the mixture is 1, that is:

1 = Xa + Xb

As Xa = 0.24

1 = 0.24 + Xb

1-0.24 = Xb

0.76 = Xb

Mole fraction of B is 0.76

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