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 $=\frac{1000\mathrm{g}}{78\mathrm{g}{\mathrm{mol}}^{-1}}=12.82 \mathrm{moles}$
Mole fraction of solute $=\frac{\mathrm{x}}{\mathrm{x}+12.82}\Rightarrow 0.2=\frac{\mathrm{x}}{\mathrm{x}+12.82}$
On solving       x = 2.
$\text{ Molality }\left(\mathrm{m}\right)=\frac{\text{ Number of moles of solute }}{\text{ Mass of solvent (in }\mathrm{kg})}=\frac{3.2}{1}=3.2$