If mole fraction of a solute in 1 kg benzene is 0.2 then molality of solute is

Let number of moles of solute in solution = x Moles of benzene in solution =1000g78gmol−1=12.82 moles Mole fraction of solute =xx+12.82⇒0.2=xx+12.82On solving       x = 2. Molality (m)= Number of moles of solute  Mass of solvent (in kg)=3.21=3.2