(a) The vapour pressures of benzene and toluene at 293 Kare 75 mm Hg and 22 mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of benzene in the vapour phase assuming that the vapour pressures are in equilibrium with the liquid mixture at this temperature.
(b) What is meant by +ve and -ve deviations from Raoult's law and how is the sign of H solution related to +ve and -ve deviations from Raoult's law ?

Explanation

(a) Calculation of the total vapour pressure Moles of benzene `(N_(A))`
` = ("Mass of Benzene")/("molar mass of benzene " (C_(6) H_(6)))`
=` ((23.4 g))/((78 "g mol"^(-1))) = 0.3 ` mol
Moles of toluene `(N_(B))`
= `(" mass of toluene")/("Molar mass of toluene " (C_(7) H_(8)))`
`= ((64.6 g))/((92g "mol"^(-1))) ` = 0.7 mol
Mole fraction of benzene `(X_(A))`
`= (n_(A))/(n_(A) + n_(B))`
= `((0.3 "mole"))/((0.3 " mole " + 0.7 "mol")) = 0.3`
M.ole fraction of toluene `(X_(B))`
`= (n_(B))/(n_(A) + n_(B))`
= `((0.7 "mole"))/((0.3 " mole" + 0.7 "mol")) = 0.7`
Vapour pressure of toluene in solution `(P_(A))`
`P_(A)^(0) xx X_(A) = (75 mm) xx 0.3`
= 22.5 mm
Vapour pressure of toluene in solution` (P_(B))`
`P_(B)^(0) xx X_(B)`
= (22 mm) `xx 0.7`
= 15.4 mm
Total vapour pressure of solution
= `P_(A) + P_(B)`
= 22.5 + 15.4
= 37.9 mm
Calculation of mole fraction of benzene in the vapour phase.
Mole fraction of benzene in vapour phase
`= (" Partial vapur pressure of benzene ")/("Total vapour pressure ")`
= `((22.5 mm))/((37.9 " mm")) = 0.591`
(b) Solutions showing positive deviations, the partial vapour pressure of each component of solution is greater than the vapour pressure as expected according to Raoult.s low.
Solutions showing negative deviations, the partial vapour pressure of each component of solution is less than vapour pressure as expected according to Raoult.s law.
For posivitve deviation `Delta H_("mixing ") = + `ve
For negative deviation `Delta_("mixing")` = -ve