Question

What is the total vapor pressure at 20°C of a liquid solution containing 0.29 mole fraction benzene, C6H6, and 0.71 mole fraction toluene, C6H5CH3? Assume that Raoult’s law holds for each component of the solution. The vapor pressure of pure benzene at 20°C is 75 mmHg; that of toluene at 20°C is 22 mmHg

Answer 1

Mole fraction of benzene,X_benzene = 0.29

mole fraction of toluene, X_toluene = 0.71

vapor pressure of pure benzene,P^o_benzene = 75 mmHg

vapor pressure of pure toluene, P^o_{toluene =} 22 mmHg

Total vapor pressure of the solution = paprtial vapor pressure of bezene + partial vapor pressure of toluene

Accorsing to Rauolts law

Partial vapor pressure of the benzene = mole fraction of the benzene * Vapor pressure of the pure benzene

Partial vapor pressure of the Toluene = mole fraction of the toluene * Vapor pressure of the pure toluene

P_solution = X_toluene * P^o_toluene + X_benzene * P^o_benzene

P_solution = 0.71 * 22 mm Hg +0.29*75 mm Hg

=37 mm Hg

Total vapor pressure of the solution = 37 mm Hg