Question

11. The boiling point at 1.236 atm pressure of a solution of 0.6589 mole fraction of benzene and 0.3411 mole fraction toluene is 88°C. At this temperature the the vapour pressures of pure benzene and toluene are 1.276 and 0.5059 atm, respectively. What is the vapour composition that boils off this liquid? [5 marks)

Answer 1

Given that:

Mole fraction of benzene= 0.6589

Mole fraction of toluene= 0.3411

Vapour pressure of pure benzene(P°_b)=1.276 atm

Vapour pressure of pure benzene(P°t)=0.5059 atm

According to the Roult law which states that the total pressure can be given by the commulative sum of partial pressure of the compents in the mixture. So,

P_total= P_b + P_t

P_total= P^°_bX_b + P^°_tX_t

P_{total= 1.276*0.6589 + 0.5059*0.3411}

P_{total= 1.0132 atm}

_{The composition in vapour phase can be given as}

Y(benzene)= P_b / P_total

_{= 0.8407/1.0132 = 0.8152}

_{while Y(toulene) will be}

_{Y(toulene)= 1- Y(benzene)}

_{~1- 0.8152= 0.1848}

^{​​​​​​_{​​​​​}}