Question

What is the freezing point of 0.251 m urea in water?

Short answer

The freezing point is$-0.{664}^{°}C$

Step-by-step solution

Concept

The freezing point of a solution containing a non-volatile solute is lower than the freezing point of a pure solvent. The depression in the freezing point can be calculated as:

${\mathbf{\Delta T}}_{\mathbf{f}}\mathbf{=}{\mathbf{iK}}_{\mathbf{f}}\mathbf{m}$

where the $K_f$is the molal freezing point depression constant and its value is equal to$1.86{}^{°}C/m$for aqueous solutions. The mis the molality of the particles, the iis the van’t Hoff factor, and its value for 1 urea is 1 because a urea is a non-electrolyte substance and it does not dissociate into ions in a solution

Calculate the depression in freezing point for urea (ΔTf)

$$\begin{gathered} ∆T_f=i{K}_{f}m \\ =1.{86}^{°}C/m\times 0.251m \\ =0.{466}^{°}C \end{gathered}$$

Freezing point of solution

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{solvent}}-{\mathrm{T}}_{\mathrm{solution}} \\ {\mathrm{T}}_{\mathrm{solvent}}=0°\mathrm{C} \\ 0°\mathrm{C}-{\mathrm{T}}_{\mathrm{solution}}=0.466°\mathrm{C} \\ {\mathrm{T}}_{\mathrm{solution}}=-0.466°\mathrm{C}. \end{gathered}$$