Pls explain this question
Solution-
In liq. state of the mix. Xbenzene = 1 /2 and Xtoluene = 1/2
acc. to Raoult"s law P total = 700 1/2 + 600 1/2 = 650 mm Hg
now partial V.P. of benzene = 350 mm Hg
the In vapour form of benzene ;
p = Xbenzene P total
Xbenzene = 350 / 650 = 0.54
( option 4)
In liq. state of the mix. Xbenzene = 1 /2 and Xtoluene = 1/2
acc. to Raoult"s law P total = 700 1/2 + 600 1/2 = 650 mm Hg
now partial V.P. of benzene = 350 mm Hg
the In vapour form of benzene ;
p = Xbenzene P total
Xbenzene = 350 / 650 = 0.54
( option 4)