--%>

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 Symmetry Types The number of

    The number of molecular orbitals and molecular motions of each symmetry type can be deduced. Let us continue to use the C2v point group and the H2O molecule to illustrate how the procedure develop

  • Q : Strength of Nacl in solution To 5.85gm

    To 5.85gm of Nacl one kg of water is added to prepare of solution. What is the strength of Nacl in this solution (mol. wt. of nacl = 58.5)? (a) 0.1 Normal (b) 0.1 Molal (c) 0.1 Molar (d) 0.1 FormalAnswer:

  • Q : Procedure for separating the components

    Briefly describe the procedure for separating the components of the gun-powder?

  • Q : Explain Second Order Rate Equations.

    Integration of the second order rate equations also produces convenient expressions for dealing with concentration time results.A reaction is classified as second order if the rate of the reaction is proportional to the square of the concentration of one o

  • Q : What are different mechanisms for

    Nucleophilic substitution reactions in halides containing  - X bond may take place through either of the two different mechanisms,S<

  • Q : Explain reactions of carbonyl oxygen

    In these reaction oxygen atom of carbonyl group is replaced by either one divalent group or two monovalent groups. Reaction by ammonia derivatives: aldehydes and ketones react with a number of ammonia derivatives such as hydroxylaminem hydrazine, semicarbazide etc. in weak acidic medium.

  • Q : Explain structure basicity of amines.

    Basic character of amines is related to their structural arrangement. Basic strength of amines depends on the relative ease of formation of the corresponding cation by accepting a proton from the acid. Greater the stability of cation is, more is basic strength of amine.Alkyl a

  • Q : Vapour pressure Vapour pressure of

    Vapour pressure of methanol in water Give me answer of this question. An aqueous solution of methanol in water has vapour pressure: (a) Equal to that of water (b) Equal to that of methanol (c) More than that of water (d) Less than that of water

  • Q : What is synthetic rubber and how it

    To meet human needs, scientists have started preparing synthetic rubbers. Besides having similar properties as natural rubbers they are tougher, more flexible and more durable than natural rubber. They are capable of getting stretched to twice its length. Though, it reverts to its original shape

  • Q : Problem based on normality Choose the

    Choose the right answer from following. NaClO solution reacts with H2SO3 as,. NaClO + H2SO3→NaCl+ H2SO4. A solution of NaClO utilized in the above reaction contained 15g of NaClO per litre. The