--%>

What is depression in freezing point?

Freezing point of a substance is the temperature at which solid and liquid phases of the substance coexist. It is defined as the temperature at which its solid and liquid phases have the same vapour pressure.

The freezing point of a pure liquid is preset. Now, if a non-volatile solute is dissolved in the pure liquid to constitute a solution, there occurs a lowering in the freezing point. The freezing point of solution refers to the temperature at which the vapour pressure of the solvent in two phases, i.e. liquid solution and solid solvent is the same. Since the vapour pressure of the solvent at a lower temperature.

Evidently the freezing point of the pure solvent is the temperature corresponding to the point B (T0 K) and that of the solution is the temperature corresponding to the point A'(T1 K). Clearly, (T0 - T1) or ΔTƒ is the freezing point depression. Since its magnitude is determined by that of lowering of vapour pressure, the freezing point depression depends upon the molal concentration of the solute and does not depend upon the nature of solid. It is, thus, a colligative property. The general relation between these two quantities for a solution of non-electrolyte is usually expressed in term of molality of the solution

ΔTƒ  Δp and Δp xB

ΔTƒ = kxB =415_freezing point.png 


For dilute solution, 272_freezing point1.png   and hence,1964_freezing point2.png.


ΔTƒ = k 1278_freezing point3.png  = k582_elevation in boiling point4.pngMA


If WA is the mass of solvent in kg, then   is equal to molality (m) of the solution

ΔTƒ = kMAm     (? kMA = Kƒ)

ΔTƒ =Kƒm, where Kƒ is called Freezing point depression constant or molal depression constant or cryoscopic constant.

As is clear from the above, depression in freezing point depends upon relative number of moles of solute and solvent but does not depend upon nature of solute, so it is a colligative property.

   Related Questions in Chemistry

  • Q : Question 6 A student was analyzing an

    A student was analyzing an unknown containing only Group IV cations. When the unknown was treated with 3M (NH4)2CO3 solution, a white precipitate formed. Because the acetic acid bottle was empty, the student used 6M HCl to dissolve the precipitate. Following the procedure of this experiment, the stu

  • Q : Normality of sulphuric acid Help me to

    Help me to go through this problem. Normality of sulphuric acid is: (a) 2N (b) 4N (c) N/2 (d) N/4

  • Q : Unit of mole fraction Provide solution

    Provide solution of this question. Unit of mole fraction is: (a) Moles/litre (b) Moles/litre2 (c) Moles-litre (d) Dimensionless

  • Q : Molecular Properties Symmetry Molecular

    Molecular orbitals and molecular motions belong to certain symmetry species of the point group of the molecule.Examples of the special ways in which vectors or functions can be affected by symmetry operations are illustrated here. All wave functions soluti

  • Q : Solubility of a gas The solubility of a

    The solubility of a gas in water depends on: (a) Nature of the gas (b) Temperature (c) Pressure of the gas (d) All of the above. Can someone help me in finding out the right answer.

  • Q : Problem related to molality Help me to

    Help me to solve this problem. What is the molality of a solution which contains 18 g of glucose (C6,H12, O6) in 250 g of water:  (a) 4.0 m (b) 0.4 m (c) 4.2 m (d) 0.8 m

  • Q : Macromolecules what are condensation

    what are condensation polymerization give in with 2 examples

  • Q : Calculating amount of Sodium hydroxide

    Choose the right answer from following. The amount of NaOH in gms in 250cm3 of a0.100M NaOH solution would be : (a) 4 gm (b) 2 gm (c) 1 gm (d) 2.5 gm

  • Q : Molar solution of sulphuric acid Choose

    Choose the right answer from following. The molar solution of sulphuric acid is equal to: (a) N solution (b) 2Nsolution (c) N/2solution (d) 3Nsolution

  • Q : Determining concentration in ppm A 500

    A 500 gm tooth paste sample has 0.2g fluoride concentration. Determine the concentration of F in terms of ppm level: (a) 250 (b) 200 (c) 400 (d) 1000Answer: (c) F-ions in ppm = (0.2/500) x 106 = 400