--%>

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 : Molecular weight of solute Select right

    Select right answer of the question. A dry air is passed through the solution, containing the 10 gm of solute and 90 gm of water and then it pass through pure water. There is the depression in weight of solution wt by 2.5 gm and in weight of pure solvent by 0.05 gm. C

  • Q : Tetrahedral holes In zinc blende

    In zinc blende structure, zinc atom fill up:(a) All octahedral holes  (b) All tetrahedral holes  (c) Half number of octahedral holes  (d) Half number of tetrahedral holesAnswer: (d) In zinc blende (ZnS

  • Q : Liquid Vapour Free Energies The free

    The free energy of a component of a liquid solution is equal to its free energy in the equilibrium vapour.Partial molal free energies let us deal with the free energy of the components of a solution. We use these free energies, or simpler concentration ter

  • Q : Molarity in Nacl The molarity of 0.006

    The molarity of 0.006 mole of NaCl in 100 solutions will be: (i) 0.6 (ii) 0.06 (iii) 0.006 (iv) 0.066 (v) None of theseChoose the right answer from above.Answer: The right answer is (ii) M = n/ v(

  • Q : Neutralisation of phosphorous acids

    Provide solution of this question. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3 PO3) the volume of 0.1 M aqueous KOH solution required is: (a) 40 mL (b) 20 mL (c) 10 mL (d) 60 mL

  • Q : Basicity order order of decreasing

    order of decreasing basicity of urea and its substituents

  • Q : Problem based on molality of glucose

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Finding strength of HCL solution Can

    Can someone please help me in getting through this problem. 1.0 gm of pure calcium carbonate was found to require 50 ml of dilute  HCL for complete reaction. The strength of the HCL  solution is given by: (a) 4 N  (b) 2 N  (c) 0.4 N  (d) 0.2 N

  • Q : Problem on volumetric flow rate Methane

    Methane containing 4 mol% N2 is flowing through a pipeline at 105.1 kpa and 22 °C. To check this flow rate, N2 at the same temperature and pressure are introduced to the pipeline at the rate of 2.83 m3/min. At the end of the pipe (

  • Q : Chem Explain how dissolving the Group

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.