--%>
Assignment Help - homework help

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 : What is Distillation Separation by

    Separation by distillation can be described with a boiling point diagram. The important process of distillation can now be investigated. From the boiling point diagram one can see that if a small amount of vapour were removed from a liquid of composit

    		•
  • Q : Symmetry Elements The symmetry of the

    The symmetry of the molecules can be described in terms of electrons of symmetry and the corresponding symmetry operations.Clearly some molecules, like H2O and CH4, are symmetric. Now w

    		•
  • Q : Alkaline medium The amount of KMnO 4

    The amount of KMnO4 required to prepare 100 ml of 0.1N solution in alkaline medium is: (a) 1.58 gm (b) 3.16 gm (c) 0.52 gm (d) 0.31 gmAnswer: (a) In alkaline medium KMnO4 act as ox

    		•
  • Q : BASIC CHARACTER OF AMINES IN GAS PHASE,

    IN GAS PHASE, BASICITIES OF THE AMINES IS JUST OPPOSITE TO BASICITY OF AMINES IN AQEUOUS PHASE .. EXPLAIN

    		•
  • Q : Question based on vapour pressure and

    Benzene and toluene form nearly ideal solutions. At 20°C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The parial vapour pressure of benzene at 20°C for a solution containing 78g of benzene and 46g of toluene in torr is: (a) 50 (b)

    		•
  • Q : Molarity based question Help me to

    Help me to solve this problem. 4.0 gm of NaOH are contained in one decilitre of solution. Its molarity would be: (a) 4 M (b)2 M (c)1 M (d)1.5 M

    		•
  • Q : Utilization of glacial acetic acid What

    What is the utilization of glacial acetic acid? Briefly describe the uses.

    		•
  • Q : Changes in matter law of chemical

    changes in matter law of chemical combination

    		•
  • Q : Chemists have not created a periodic

    Explain the reason behind that the chemists have not created a periodic table of compounds?

    		•
  • Q : Unit of molality Select the right

    Select the right answer of the question. The unit of molality is: (a) Mole per litre (b) Mole per kilogram (c) Per mole per litre (d) Mole litre

    		•