Freezing point depression refers to the lowering of the freezing point of solvents upon the addition of solutes. It is a colligative property of solutions that is generally proportional to the molality of the added solute. The depression in the freezing point of a solution can be described by the following formula. ΔTf = i×Kf×m Where Freezing Point Depression in Solutions As per Raoult’s law, “the vapour pressure of a pure solvent decreases with the addition of a solute”. Since the vapour pressure of a non-volatile solvent is zero, the overall vapour pressure of the solution is lesser than that of the pure solvent.
Recommended Videos
Why does the Freezing Point Depression Occur?
The reason for the depression of the freezing point of a solvent upon the addition of a solute is explained below.
- At the freezing point of a solvent, there exists an equilibrium between the liquid state and the solid state of the solvent.
- This implies that the vapour pressures of both the liquid and the solid phase are equal.
- Upon the addition of a solute which is non-volatile, the vapour pressure of the solution is found to be lower than the vapour pressure of the pure solvent.
- This causes the solid and the solution to reach equilibrium at lower temperatures.
A graph detailing the freezing point depression of water upon the addition of sucrose to it is provided below.
Freezing Point Depression Graph
From the graph, it can be observed that the increase in the molality of sucrose causes further depression in the freezing point of the solvent (water).
Freezing Point Examples
- The freezing point of seawater is below 0oC; it remains liquid at temperatures below the freezing point of pure water. This is caused by the salts that are dissolved in it.
- Another example of freezing point depression of a solvent can be observed in vodka. It can be considered to be a solution of ethanol in water, and its freezing point is lower than water but higher than pure ethanol.
- Many organisms can survive in freezing climates because their bodies can produce compounds such as glycerol and sorbitol, which helps in decreasing the freezing point of the water in their bodies.
The normal freezing point and the corresponding freezing point depression is tabulated below.
| Solvent | Normal Freezing Point, oC | Freezing Point Depression, Kb, oC m-1 |
| Water | 0.0 | 1.86 |
| Acetic acid | 16.6 | 3.9 |
| Benzene | 5.5 | 5.12 |
| Chloroform | -63.5 | 4.68 |
| Nitrobenzene | 5.67 | 8.1 |
Uses of Freezing Point Depression
Some important uses of freezing point depression are listed below.
- In cold areas where the temperatures range from 0oC to -15oC, sodium chloride is spread over the roads in order to lower the freezing point of water and prevent the build up of ice.
- If the temperatures are below 18oC, calcium chloride is used instead of NaCl to melt the ice on the roads. This is because CaCl2 dissociates into 3 ions, causing a greater depression in the freezing point of water.
- Radiator fluids used in many automobiles are generally made up of ethylene glycol and water. This prevents the freezing of the radiator during cold seasons.
- The molar mass of a given solute can be determined from the freezing point depression formula.
- The degree to which a solute dissociates in a solvent can also be measured with the help of this formula.
Freezing point, when a liquid becomes a solid. Increased pressure, as with the melting point, typically increases the freezing point. Bringing a seed crystal into a supercooled liquid causes freezing, resulting in the release of fusion heat increasing the temperature to the freezing point rapidly.
Liquids have a temperature characteristic at which they become solids, known as their freezing point. Theoretically, a solid’s melting point should be the same as the liquid’s freezing point. During the action, it is possible to observe small differences between these quantities.
Fusion, vaporization and sublimation are endothermic processes, while exothermic processes are freezing, condensation, and deposition.
Water freezes when it hits 32 degrees Fahrenheit (0 degrees Celsius), but the amount of time it takes to do so depends on several variables that may vary from your neighbour’s in your freezer.
The six-phase changes are freezing, boiling, condensing, vaporizing, sublimating, and deposition.
To learn more about this property of solutions and other colligative properties, such as the elevation in boiling point, register with BYJU’S and download the mobile application on your smartphone.
Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!
Select the correct answer and click on the “Finish” button
Check your score and answers at the end of the quiz
Visit BYJU’S for all Chemistry related queries and study materials
0 out of 0 arewrong
0 out of 0 are correct
0 out of 0 are Unattempted
View Quiz Answers and AnalysisAnswer
Hint: By using the formula of depression in freezing point one can determine the relation between the freezing point and the molecular weight. The depression in freezing point is equal to the product of cryoscopic constant and molality.
Complete step by step answer:
The freezing point depression is explained as the lowering of the freezing point of the solvent with the addition of the solute.The freezing point depression is the colligative property which is proportional to the molality of the solute added.The freezing point depression is given by the formula as shown below.$\Delta {T_f} = {K_f} \times m$Where,$\Delta {T_f}$ is the freezing point depression${K_f}$ is the cryoscopic constantm is the molalityThe molality is defined as the number of moles of solutes dissolved in one kilogram of solvent.The formula of molality is shown below.$m = \dfrac{n}{{{M_1}}}$Where,m is the molalityn is the number of moles${M_1}$ is the mass in kilogramThe number of moles is given by the formula as shown below.$n = \dfrac{m}{{{M_2}}}$Where,n is the number of molesm is the mass${M_2}$ is the molecular weightSo, if we substitute the terms in the formula of freezing point depression then the new formula obtained will be.$ \Rightarrow \Delta {T_f} = {K_f} \times \dfrac{m}{{{M_2}{M_1}}}$If we keep the ${K_f}$, m and ${M_1}$ as constant then the freezing point depression will be inversely proportional to the molecular weight.$ \Rightarrow \Delta {T_f} = \dfrac{1}{{{M_2}}}$So, by increasing the value of depression in freezing point, the molecular weight decreases and vice versa.Thus, an increase in the molecular weight will have a smaller effect on the freezing point.
Note:
The freezing point is calculated by subtracting the depression in freezing point value and the freezing point of pure water which is zero degree Celsius.