Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 g of urea per 250 gm of water (Mol. wt. of urea = 60).

Weight of solute (urea) dissolved = 3.0 g

Weight of the solvent (water) = 250 g

Molecular weight of the solute = 60

3.0 gm of the solute $= \frac{3.0}{60}$ moles 

                                            = 0.05 moles 

$\mathrm{Molality} = \frac{\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute} \mathrm{x} 1000 \mathrm{g} \mathrm{of} \mathrm{solvent} }{\mathrm{Weight} \mathrm{of} \mathrm{solvent}}$

$\therefore$ Molality $=\frac{3/60}{250}\times 1000=0.2\mathrm{m}$

Calculation of mole fraction 

3.0 g of solute = 3/60 moles = 0.05 mole 

250 g of water $=\frac{250}{18}$ moles = 13.94 moles 

$\therefore$ Mole fraction of the solute $=\frac{0.05}{0.05+13.94}=\frac{0.05}{13.99}=0.00357$