Mole fraction of a solute in benzene is 0.2. The molality of solute is

Let the solution be x molal then moles of benzene present in 1000 g of benzen =100078=12.82mol∴Mole fraction of solute =xx+12.82                                 0.2=xx+12.820.2(x+12.82)=x   0.2x+2.564=x               2.564=x−0.2x′=0.8x                       x=2.85640.8=3.57