Solve this:
Q. Equal moles of benzene and toluene are mixed the V.P. of benzene and toluene in pure state are 700 and 600 mm Hg respectively. The mole fraction of benzene in vapour state is
(1) 0.7
(2) 0.47
(3) 0.50
(4) 0.54

Dear student ,
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)
Regards

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