Find the freezing point of a solution containing 27\ g urea in 1\...

Number of moles of urea $= \dfrac{\text{Mass}}{\text{Molar mass}} = \dfrac{27}{60} = 0.45\ \text{mol}$

Given the volume of the aqueous solution is $1\ \text{dm}^3$ ($1\ \text{L}$). For a dilute aqueous solution, we can approximate the mass of the solvent to be equal to the mass of $1\ \text{L}$ of water, which is $1\ \text{kg}$.

Molality ($m$) $= \dfrac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \dfrac{0.45}{1} = 0.45\ \text{mol kg}^{-1}$

Depression in freezing point is given by:

$\Delta T_f = K_f \times m$

$\Delta T_f = 1.8 \times 0.45 = 0.81\ \text{K}$

The freezing point of pure water is 0\ ^{}^{\circ}\text{C}. Therefore, the freezing point of the solution is:

T_f = 0 - \Delta T_f = -0.81\ ^{}^{\circ}\text{C}

Answer: -0.81\ ^{}^{\circ}\text{C}