FREEZING POINT,CRYOSCOPIC CONSTANT

1. Colligative Property (freezing point depression)
∆𝑻 𝒇 = 𝑲 𝒇 m ←molality which is the amount of solute added
↑ ↑
↑ constant which is the freezing point elevation constant
Temperature
decrease because of solute
For water, 𝐾𝑏 = 1.86 °𝐶/molal
So the boiling point elevation is ∆𝑻 𝒇 = −𝑲 𝒇 𝒎 = - (1.86)(1.25) = 2.32℃
And the new f.p. of water with a heck of a lot of sugar in it is -2.32 ℃
Depression of Freezing Point
The lowering of vapour pressure of a solution causes a lowering of the freezing point compared to
that of the pure solvent.
Examples
Examples include salt in water, alcohol in water, or the mixing of two solids such as impurities in
a finely powdered drug.
Explanation:
We know that at the freezing point of a substance, the solid phase is in dynamic equilibrium with
the liquid phase. Thus, the freezing point of a substance may be defined as the temperature at
which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the
solid phase. A solution will freeze when its vapour pressure equals the vapour pressure of the pure
solid solvent. According to Raoult’s law, when a non-volatile solid is added to the solvent its
vapour pressure decreases and now it would become equal to that of solid solvent at lower
temperature. Thus, the freezing point of the solvent decreases.

2. Fig. Diagram showing ∆𝑇𝑓 depression of freezing point of a solvent in solution
Let 𝑇𝑓 𝑜
be the freezing point of pure solvent and Tf be its freezing point when non-volatile solute
is dissolved in it.
The decrease in freezing point.
∆Tf = − To -Tf is known as depression in freezing point.
Similar to elevation of boiling point, depression of freezing point (∆Tf ) for dilute solution
(idealsolution) is directly proportional to molality, m of the solution. Thus,
∆Tf ∝ m
∆Tf = Kf m
The proportionality constant, Kf , which depends on the nature of the solvent is known as Freezing
Point Depression Constant or Molal Depression Constant or Cryoscopic Constant. The unit of Kf
is K kg mol-1. Values of Kf for some common solvents are listed in Table. If w2 gram of the solute
having molar mass as M2 , present in w1 gram of solvent, produces the depression in freezing point
∆Tf of the solvent then molality of the solute
𝒎 =
𝒘 𝟐/𝑴 𝟐
𝒘 𝟏/𝟏𝟎𝟎𝟎
Substituting this value of molality in equation we get:
∆𝑻 𝒇 =
𝑲 𝒇 × 𝒘 𝟐 × 𝟏𝟎𝟎𝟎
𝒘 𝟏/𝟏𝟎𝟎𝟎
∆𝑻 𝒇 =
𝑲 𝒇 × 𝒘 𝟐 × 𝟏𝟎𝟎𝟎
𝑴 𝟐 × 𝒘 𝟏
𝑴 𝟐 =
𝑲 𝒇 × 𝒘 𝟐 × 𝟏𝟎𝟎𝟎
∆𝑻 𝒇 × 𝒘 𝟏
Thus for determining the molar mass of the solute we should know the quantities w1 , w2 , ∆Tf ,
along with the molal freezing point depression constant. The values of Kf and Kb, which depend
upon the nature of the solvent, can be ascertained from the following relations.

3. 𝑲 𝒇 =
𝑹 × 𝑴 𝟏 × 𝑻𝒇 𝟐
𝟏𝟎𝟎𝟎 × ∆ 𝒇𝒖𝒔 𝑯
𝑲 𝒃 =
𝑹 × 𝑴 𝟏 × 𝑻𝒃 𝟐
𝟏𝟎𝟎𝟎 × ∆ 𝒗𝒂𝒑 𝑯
Here the symbols R and M1 stand for the gas constant and molar mass of the solvent, respectively
and Tf and Tb denote the freezing point and the boiling point of the pure solvent respectively in
kelvin. Further, ∆fusH and ∆vapH represent the enthalpies for the fusion and vaporization of the
solvent, respectively.
Freezing point depression constants for some solvents
The molal freezing point depression constant, Kf, has a specific value depending on the identity of
the solvent.:
solvent
Normal
freezing
point, oC
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
Determination of the molecular mass
Generally freezing point depression is used to determine the molecular mass of an unknown
𝒎 =
∆𝑻 𝒇
𝑲 𝒇
=
𝒎𝒐𝒍𝒆𝒔
𝑲𝒈 𝒔𝒐𝒍𝒗𝒆𝒏𝒕
𝑴. 𝑾.=
𝒎𝒂𝒔𝒔 𝒔𝒐𝒍𝒖𝒕𝒆
𝒎𝒐𝒍𝒂𝒍. 𝑲𝒈
=
𝒈 𝒔𝒐𝒍𝒖𝒕𝒆
𝒎𝒐𝒍𝒆 𝒔𝒐𝒍𝒖𝒕𝒆

4. Cryoscopic apparatus
For cryoscopy, the apparatus to measure freezing point depression of a pure solvent may be
representative of the Beckmann apparatus .The apparatus consists of a test tube containing the
solute dissolved in a pure solvent, stir bar or magnetic wire and closed with a rubber stopper
encasing a mercury thermometer. The test tube component is immersed in an ice-water bath in a
beaker. The rubber stopper and stir bar/wire stirrer.
Cryoscopic Method
The detailed information about the procedure used for cryoscopy is shown below:
1. Weigh (15 to 20 grams) of the pure solvent in a test tube and record the measured weight
value of the pure solvent.
2. Place a stir bar or wire stirrer in the test tube and close with a rubber stopper that has a hole
to encase a mercury thermometer.
3. Place a mercury thermometer in the rubber stopper hole.
4. Immerse the test tube apparatus in an ice-water bath.
5. Allow the solvent to stir continuously and equilibrate to a few degrees below the freezing
point of the solvent.
6. Record the temperature at which the solvent reaches the freezing point, which remains at a
constant temperature reading.
7. Repeat the freezing point data collection for at least two more measurements without a
difference less than 0.5 °C between the measurements.
8. Weigh a quantity of the solute for investigation and record the measured value.
9. Add the weighed solute to the test tube containing the pure solvent.
10. Re - close the test tube with a rubber stopper encasing a mercury thermometer.
11. Re-immerse the test tube in an ice water bath and allow the mixture to stir to fully dissolve
the solute in the pure solvent.
12. Measure the freezing point and record the temperature value.
Allow the solution to stir continuously to avoid supercooling.
The observed freezing point of the solution is when the temperature reading remains constant

5. Fig. Beckmann’s Apparatus
Uses:
 The use of salt to de-ice roads is a common application of this principle. The solution
formed when some of the salt dissolves in the moist ice reduces the freezing point of the
ice. If the freezing point falls below the ambient temperature, the ice melts. In very cold
weather, the ambient temperature may be below that of the salt solution, and the salt will
have no effect.
 Automotive radiator antifreezes are mostly based on ethylene glycol, (CH2OH).Owing
to the strong hydrogen-bonding properties of this double alcohol, this substance is miscible
with water in all proportions, and contributes only a very small vapor pressure of its own.
Besides lowering the freezing point, antifreeze also raises the boiling point, increasing the
operating range of the cooling system.

6. References:
 Knovel Critical Tables, 2nd Ed., Knovel, New York (2008).
 http://www.lahc.cc.ca.us/classes/chemistry/arias/Exp%2012%20
 %20Freezing%20Point.pdf
 H. C. Jones, The Modern Theory of Solutions: Memoirs by Pfeffer, Van't Hoff, Arrhenius,
and Raoult, Harper and Brothers, New York (1899).
 D. A McQuarrie, P. A Rock, and E. B. Gallogly, General Chemistry, University Science
Books, Mill Valley (2011)
 Lecture of Dr.Faizan