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Chemistry Class 12

Class 12 : Doc: Colligative Properties of Solutions Notes | EduRev

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What are Colligative Properties? 

  • A dilute solution is one in which the amount of the solute is very small in comparison to the amount of the solvent. 
  • The dilute solutions show more or less ideal behaviour as the heat and volume changes, accompanying the mixing of solute and solvent, are negligible for all practical purposes. Dilute solutions obey Raoult’s law. 
  • The properties of dilute solutions which depend only on the number of particles of solute present in the solution and not on their identity are called colligative properties (denoting depending upon collection).
  • We shall assume here that the solute is non-volatile, so it does not contribute to the vapour. We shall also assume that the solute does not dissolve in the solid solvent, that is, the pure solid solvent separates when the solution is frozen, the latter assumption is quite drastic, although it is true for many mixtures, it can be avoided.

Colligative properties are the properties of only dilute solution which are supposed to behave as ideal solutions. 

Different Types of Colligative Properties of Solution

The various colligative properties are:

  • Lowering of vapour pressure
  • Elevation of boiling point
  • Depression of freezing point
  • Osmotic pressure

Doc: Colligative Properties of Solutions Notes | EduRev

Try yourself:Which of the following is a colligative property?
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Relative Lowering of Vapour Pressure

  • It has been known for a long time that when a non-volatile solute is dissolved in a liquid, the vapour pressure of the solution becomes lower than the vapour pressure of the pure solvent. 
  • In 1886, the French chemist, Francois Raoult, after a series of experiments on a number of solvents including water, benzene and ether, succeeded in establishing a relationship between the lowering of the vapour pressure of a solution and the mole fraction of the non-volatile solute.
  • Let us consider a solution obtained by dissolving n moles of a non-volatile solute in N moles of a volatile solvent. Then mole fraction of the solvent, X1 = N/(n + N) and mole fraction of the solute, X2 = n /(N + n). Since the solute is non-volatile, it would have a negligible vapour pressure. The vapour pressure of the solution is, therefore merely the vapour pressure of the solvent. 

According to Raoult's law, the vapour pressure of a solvent (P1) in an ideal solution is given by the expression:
P1 = X1 Pº1      ...(1)
where Pº1 is the vapour pressure of the pure solvent. Since X1 + X2 = 1, Eq. 1 may be written as
P1 = (1 - X2) Pº1      ...(2)
or P1/Pº1 = 1 - X2
or P1 - Pº1/Pº1 = X2    ...(3)
The expression on the left-hand side of Equation (3) is usually called the relative lowering of vapour pressure. Equation (3) may thus be stated as: 

The relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute present in the solution.

This is one of the statements of Raoult's law.
Since mole fraction of the solute, X2 is given by n/(N + n), Equation (3) may be expressed as Pº1 - P1/Pº1 = n/N + n  ....(4)
It is evident from Equation (4) that the lowering of the vapour pressure of a solution depends upon the number of moles (and hence on the number of molecules) of the solute and not upon the nature of the solute dissolved in a given amount of the solvent. Hence, lowering vapour pressure is a colligative property.

Determination of Molar Masses from Lowering of Vapour Pressure

  • It is possible to calculate molar masses of non-volatile non-electrolytic solutes by measuring the vapour pressures of their dilute solutions.
  • Suppose, a given mass, w gram, of a solute of molar mass m, dissolved in
    W gram of solvent of molar mass M lowers the vapour pressure from Pº1 to P1.

Then, by Equation  (4)
Doc: Colligative Properties of Solutions Notes | EduRev

Try yourself:Which law specifically governs the relative lowering of vapor pressures in solutions?
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Measurement of Lowering of Vapour Pressure

  1. Barometric Method: 
    The individual vapour pressure of a liquid was calculated by Raoult and then the same process was followed to calculate the vapour pressure of the solution as well. He poured the liquid or the solution into a Torricellian vacuum of a barometer tube and calculated the depression of the mercury level. This method was later found to be neither practicable nor accurate as the lowering of vapour pressure is almost negligible.
  2. Manometric Method: 
    The vapour pressure of a liquid or solution can be fairly measured with the help of a manometer. Let us assume a bulb is charged with the liquid or solution. The air in the connecting tube of the instrument is then removed with a vacuum pump. With the stopcock being closed, the pressure inside is only due to the vapour evaporating from the solution or liquid. This method can be applied to aqueous solutions. The manometric liquid used can be either mercury or n-butyl phthalate which has a low density and low volatility.
  3. Ostwald and Walker’s Dynamic Method (Gas Saturation Method):
    When air is passed through any gas, the gas diffuses into the air due to the principle of diffusion till the pressure of the gas in the air is equal to that of the gas outside. Therefore as air is passed through the solution, it absorbs the vapours of the solvent till the pressure of the solvent vapours in the air is P (vapour pressure of the solvent in the solution).
    Mass of vapour absorbed or loss in weight of solution = PVm/RT
    Where, P = pressure of the solvent vapours,
    V = volume available for vapours in the solution bulb,
    T = temperature of the bulb and
    m = molecular weight of the solvent.
    As the air passes through the solvent, the pressure difference of the vapours in the air and outside is Pº - P. So the air absorbs this amount of vapours and gets saturated.
    Loss in weight of solvent bulb = (P° - P)Vm/RT
    As the sum of the losses of weight in the two containers is equal to the gain in weight of CaCl2,
    The gain in weight of CaCl2 filled tubes =
    Doc: Colligative Properties of Solutions Notes | EduRevMethod: In this method, a stream of dry air is passed successively through
    (i) a solution
    (ii) the pure solvent (water) and
    (iii) a reagent (anhydrous CaCl2) that can absorb the vapours of the solvent. The complete assembly is shown in the figure, given below:

Doc: Colligative Properties of Solutions Notes | EduRev

Ostwald - Walker Experiment
  • The first three bulbs contain a weighed amount of the solution under examination and the next three bulbs contain a weighed amount of the pure solvent.
  • A weighed amount of CaCl2 is taken in the set of U - tubes at the end. All the bulbs are at the same temperature and the volume available for the vapours in the solution and solvent bulbs are the same temperatures and the volume available for the vapours in the solution and solvent bulbs are the same.
It is designed in order to calculate the relative lowering of vapour pressure of a solvent due to a non - volatile solute.

Practice Question: The density of a 0.438 M solution of potassium chromate at 298 K is 1.063 g cm-3. Calculate the vapour pressure of water above this solution. Given : Pº (water) = 23.79 mm Hg.
Solution. 

A solution of 0.438 M means 0.438 mol of K2CrO4 is present in 1L of the solution. Now,
Mass of K2CrO4 dissolved per litre of the solution  = 0.438 × 194  = 84.972 g
Mass of 1L of solution = 1000 × 1.063 = 1063 g
Amount of water in 1L of solution = 978.028/18  = 54.255 mol
Assuming K2CrO4 to be completely dissociated in the solution, we will have;
Amount of total solute species in the solution = 3 × 0.438 = 1.314 mol.
Mole fraction of water solution = 54.335/(54.335 + 1.314) = 0.976
Finally, Vapour pressure of water above solution = 0.976 × 23.79  = 23.22 mm Hg

Try yourself: If the relative lowering of pressure of o-xylene is 0.005 due to addition of 0.5 grams of non-volatile solute in 500 grams of solvent, what is the molecular weight of the solute?
View Solution

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