Question

A solution of benzene (C6H6) and toluene (C7H8) is 25.0 % benzene by mass. The vapor pressures of pure benzene and pure toluene at 25∘C are 94.2 and 28.4 torr, respectively.

A)Assuming ideal behavior, calculate the vapor pressure of benzene in the mixture.

B)Assuming ideal behavior, calculate the vapor pressure of toluene in the mixture.

Answer 1

P_b = P^o_b X x_b

P_b = Partial pressure of benzene

P^o_b = Vapour Presure of pure bezene

x_b = Mole fration of benzene

x_b = moles of benzene / (moles of benzene + toluene)

moles of benzene = mass of benzene / molar mass of benzne

mass of benzene = 25g. As the solution is 25% bezene by mass, it means 25g of benzene in 100g of solution.

moles of benzene = 25 / 72 = 0.35 moles

% of toluene = 75%

Moles of toluene = 75 / 86 = 0.87 moles

Mole fraction of benzene = 0.35 / (0.35 + 0.87) = 0.29

Mole fraction of toluene = 0.87 / (0.87 + 0.35) = 0.71

P_b = 94.2 X 0.29 = 27.32 torr

And

P_t = 28.4 X 0.71 = 20.161 torr