An X molal solution of a compound in benzene has mole fraction of solute = 0.2. The value of X is

Mole fraction of solute X2=0.2. Therefore, mole fraction of solvent X1=0.8Or n2n1+n2=0.2 and n1n1+n2=0.8Or n2n1=0.20.8=14Now, if n1 (solvent moles) = 1000/78 = 12.8 moles n2=12.8/4=3.2 moles. Therefore, 3.2 moles of the compound are present in one kg of solvent benzene and so molality = 3.2.                                                                                   ORMole fraction of solute =  mm+1000Molecular weight of solvent                           0.2   = mm+100078                     ⇒0.2 = mm+12.82                       m = 0.2m+ 12.82×0.2                       m = 3.2