Question

Calculate the molality and mole fraction of solute in aqueous solution containing 3g of urea in 250g of water .

Answer 1

molality = 0.2
mole fraction of solute = 0.00358

Answer 2

Answer:

The molality and mole fraction of solute in aqueous solution containing 3g of urea in 250g of water is  0.2 mol/kg and 0.05 respectively.

Explanation:

Mass of urea in solution = 3 g

moles of urea = $n_u=\frac{3 g}{60 g/mol}=0.05$

Mass of water = 250 g

Moles of water =$n_w=\frac{250 g}{18 g/mol}=13.88mol$

Mole fraction of urea that is solute:

$\chi_u=\frac{n_u}{n_u+n_w}=\frac{0.05}{0.05+13.88}=0.0035$

$molality=\frac{\text{moles of solute}}{\text{Mass of solvent (kg)}}$

$m=\frac{0.05}{0.250 kg}=0.2 mol/kg$

The molality and mole fraction of solute in aqueous solution containing 3g of urea in 250g of water is  0.2 mol/kg and 0.05 respectively.