question_answer 50)

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressures of pure benzene and toluene at 300 K are 50.71 nun Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Answer:

Molar mass of benzene (C6H6) = 78 g mol?1 Molar mass of toluene (C6H5CH3) = 92 g mol?1 No. of moles in 80 g of benzene (nbenzene) No. of moles in 100 g of toluene (ntoluene) No. of moles in 100 g of toluene (ntoluene)            In the solution, mole fraction of benzene = 0.486 Mole fraction of toluene  = 50.71 mm, = 32.06 mm Applying Raoult?s law, = 0.486 × 50.71 mm = 24.65 mm = 0.514 × 32.06 mm = 16.48 mm Mole fraction of benzene in the vapour phase