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.

Benzene (C_6 H_6) and toluene (C_7 H_8) form a nearly ideal solution.

1.	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.
Answer» Solution : 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 partialvapourpressureof benxzene`(P_A ) =P_(A)^(@)X_A` `=150 XX 0.541 = 81.15m m` Particalvapourpressureof toluene `(P_B) =P_(B)^(@)X_B` `=50 xx 0.459 =22.95 m m` totalvapourpressureof MIXTURE=`81.15 +22.95= 104.1 m m`