Question

A toluene solution is blended with 25% mass of benzene. The vapor pressure of benzene and toluene at 25'C are 94.2 torr and 28.4 torr respectively. (a) Calculate the vapor pressure of each component, (b) total pressure of the above solution, and (c) the composition of the vapor in mass percent.

Answer 1

Assume that the total mass of mixture is 100 g thus the mass of benzene is 25 g and the mass of toluene is 75 g

Number of moles of benzene = amount in g/ molar mass

= 25 g/ 78.11 g/ mole

= 0.32 moles

Number of moles of toluene = amount in g/ molar mass

= 75 g/ 92.14 g/ mole

= 0.82 moles

Total number of moles = 0.32+0.82= 1.14

Mole fraction of benzene = 0.32/1.14= 0.28

Mole fraction of toluene = 0.82/1.14= 0.72

a) Calculate the vapor pressure of each component

vapor pressure of benzene = 94.2 torr * mole fraction

=94.2 torr * 0.28

= 26.38 torr

vapor pressure of toluene = 28.4 torr * mole fraction

=28.4 torr * 0.72

= 20.448 torr

(b) total pressure of the above solution=

vapor pressure of benzene + vapor pressure of toluene

= 26.38 torr +20.448 torr

= 46.828 torr

(c) the composition of the vapor in mass percent.

The number of moles of each gas in the vapor phase is proportional to their partial pressures, so:
(26.38 torr torr C6H6) x (78.11 g/mol) = 2060.54 torr·g/mol
(20.448 torr C7H8 ) x (92.14 g C8H18/mol) = 1884.08 torr·g/mol

(2060.54 torr·g/mol) / (2060.54 torr·g/mol + 1884.08 torr·g/mol)

= 2060.54 torr·g/mol) / (3944.62 torr·g/mol

=0.522 = 52.2% C6H6 by mass
100% - 52.2% = 47.8% C7H8 by mass