Q98P - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

Answer

The freezing point is $-0.{664}^{°}C$

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 m is the molality of the particles, the i is 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 ( Δ T f )

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

Freezing point of solution

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