Question

An ideal solution is made from 5.00 mol of benene and 3.25 mol of toluene. At 298K, the vapor pressure of the pure substances are p*benzene= 96.4 torr and p*toluene= 28.9 torr.

a) The pressure above this solution is reduced from normal pressure. At what pressure does the solution first boil at this temperature?

b) What is the composition (% benzene and % toluene) of the vapor under these conditions?

Answer 1

mole of benzene= 5moles. molles of toluene= 3.25 moles

mole fraction = moles of component/total moles

Mole fraction : Benzene = 5/(5+3.25)= 0.61, toluene =1-0.61=0.39

since the solutioni is ideal, Raoults law is obeyed, let x1, x2= mole fraction of benzene and toluene in the liquid phase, y1,y2= mole fractions of benzene and toluene in the vapor phase, P1sat= pure component vapor pressure of Benzene , P2dat= Pure component vapor pressure of toluene, P= total pressure

y1P=x1P1sat and y2P=x2P2sat

or y1P+ y2P= x1P1sat+x2P2sat, P= x1P1sat+x2P2sat= 0.61*96.4+0.39*28.9 Torr=69.8 torr

y1= x1P1sat/P= 0.61*96.4/69.8=0.84, y2= 1-0.84=0.16