Question

How do you determine freezing point depression?

Answer 1

Freezing point depression-

The freezing point is a property observed in a solution that results from the introduction of the solute molecules to a solvent. The freezing point of the solution is lower than that of pure solvent and is directly proportional to the molarity of the solute.

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)-{\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right) \\ ={\mathrm{K}}_{\mathrm{f}}-\mathrm{m} \\ \mathrm{where}, \\ ∆{\mathrm{T}}_{\mathrm{f}}=\mathrm{Freezing}\mathrm{point}\mathrm{of}\mathrm{depression} \\ {\mathrm{T}}_{\mathrm{f}}=\mathrm{Freezing}\mathrm{point}\mathrm{of}\mathrm{solution} \\ {\mathrm{T}}_{\mathrm{f}}=\mathrm{Freezing}\mathrm{point}\mathrm{of}\mathrm{solvent} \\ {\mathrm{K}}_{\mathrm{f}}=\mathrm{Freezing}\mathrm{point}\mathrm{depression}\mathrm{constant} \\ \mathrm{m}=\mathrm{Molality} \end{gathered}$$

Question: $1.60\mathrm{g}$ naphthalene solution was dissolved in $20\mathrm{g}$ of benzene. Given that the freezing point of pure benzene is $5.5°\mathrm{C}$, and the freezing point of the mixture is $2.8°\mathrm{C}$. What is the molal freezing point depression constant, ${\mathrm{K}}_{\mathrm{f}}$ of benzene?

Solution:

Step 1:

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solvent}\right)-{\mathrm{T}}_{\mathrm{f}}\left(\mathrm{solution}\right) \\ ∆{\mathrm{T}}_{\mathrm{f}}=(5.5°\mathrm{C}-2.8°\mathrm{C}) \\ ∆{\mathrm{T}}_{\mathrm{f}}=2.7°\mathrm{C} \end{gathered}$$

Step 2:

Molality $=$ moles of solute$/\mathrm{kg}$ of solvent

moles of naphthalene $$\begin{gathered} =(1.60\mathrm{g})(1\mathrm{mol}/128\mathrm{g}) \\ =0.0125\mathrm{mol} \end{gathered}$$

molality of solution $$\begin{gathered} =(0.0125\mathrm{mol})/(0.0200\mathrm{kg}) \\ =0.625\mathrm{m} \end{gathered}$$

Step 3:

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}=\left({\mathrm{K}}_{\mathrm{f}}\right)\left(\mathrm{m}\right) \\ (2.7°\mathrm{C})=\left({\mathrm{K}}_{\mathrm{f}}\right)(0.625\mathrm{m}) \\ {\mathrm{K}}_{\mathrm{f}}=4.3°\mathrm{C}/\mathrm{m} \end{gathered}$$

Therefore, the molal freezing point depression constant is $4.3°\mathrm{C}/\mathrm{m}$

Hence, the above explanation and numerical clearly explain the freezing point depression.