Question

An aqueous solution of urea containing 18 g urea in 1500 cm 3of solution has a density of 1.502 g / cm 3

Answer 1

Mass of the solution = $1.502 \frac{g}{cm^{3}}*1500cm^{3}$

= 2253 g

Mass of solvent = 2253 g - 18 g = 2253 g solvent $* \frac{1 kg}{1000 g} =2.253 kg$

Moles of urea = $18 g *\frac{ 1 mol}{60.06 g} = 0.300 mol$

Molality = [tex] \frac{0.300 mol}{2.253 kg} =0.133 m