Benzene `(C_6 H_6)` and toluene `(C_7 H_8)` form a nearly ideal solution. At 313 K, the vapour pressure of pure benzene is 150 mm Hg and of pure toluene is 50 mm Hg. Calculate the vapour pressure of a mixture of these two containing their equal masses at 313 K.

Explanation

Let the mass of both benzene and toluene=Wg
` therefore `Moles of benzene`= (W )/(78) ,`Moles of toluene `=(W)/(92)`
Mole fraction of benzene` =((w)/(78))/((w/(78)+W/(92)))=0.541`
Mole fraction of toluene = 1-0-541 = 0.459
partial vapour pressure of benxzene `(P_A ) =P_(A)^(@)X_A`
`=150 xx 0.541 = 81.15m m`
Partical vapour pressure of toluene `(P_B) =P_(B)^(@) X_B`
`=50 xx 0.459 =22.95 m m`
total vapour pressure of mixture =`81.15 +22.95= 104.1 m m`