 # Colligative Properties

A colligative property is one of the properties of a solution. It is applied only to solutions and it is usually dependent on the concentration or the ratio of the number of particles of the substances (solute and solvent) in a solution. Colligative property does not depend on the mass or the identity (nature) of the solute as in non-colligative property.

The word”colligative” has been adapted or taken from the Latin word “colligatus” which translates to “bound together”. In the context of defining a solution, colligative property helps us to understand how the solvent’s properties are linked to the concentration of solute in a solution.

## What are Colligative Properties?

Dilute solution containing non-volatile solute exhibit some properties which depend only on the number of solute particles present and not on the type of solute present. These properties are called colligative properties. These properties are mostly seen in dilute solutions.

We can further consider colligative properties as those properties that are obtained by the dissolution of a non-volatile solute in a volatile solvent. Generally, the solvent properties are changed by the solute where its particles remove some of the solvent molecules in the liquid phase. This also results in the reduction of the concentration of the solvent.

Meanwhile, when we talk about the given solute-solvent mass ratio, colligative properties are said to be inversely proportional to the solute molar mass.

## Colligative Properties Examples

We can observe the colligative properties of solutions by going through the following examples. If we add a pinch of salt to a glass full of water its freezing temperature is lowered considerably than the normal temperature. Alternatively, its boiling temperature is also increased and the solution will have a lower vapour pressure. There are changes in its osmotic pressure as well.

Similarly, if we add alcohol to water, the solution’s freezing point goes down below the normal temperature that is observed for either pure water or alcohol.

## Different Types of Colligative Properties of Solution

There are different types of colligative properties of a solution. These include, vapour pressure lowering, boiling point elevation, freezing point depression and osmotic pressure.

### Lowering of Vapour Pressure

In a pure solvent, the entire surface is occupied by the molecules of the solvent. If a non- volatile solute is added to the solvent, the surface now has both solute and solvent molecules; thereby fraction of surface covered by solvent molecules gets reduced. Since the vapour pressure of the solution is solely due to solvent alone, at the same temperature the vapour pressure of the solution is found to be lower than that of the pure solvent. If P0 is the vapour pressure of pure solvent and Ps is the vapour pressure of the solution. The difference Po – Ps is termed as lowering in vapour pressure. The ratio Po – Ps / Po is known as the relative lowering of vapour pressure.

Raoult, in 1886, established a relation between relative lowering in vapour pressure and mole fraction. The relationship is known as Raoult’s law. It states that the relative lowering in vapour pressure of a dilute solution is equal to the mole fraction of the solute present in the solution

If n moles of solute is dissolved in N moles of the solvent, then according to Raoult’s law

Po – Ps / Po = n / n + N

### Elevation in Boiling Point

The boiling point of a liquid is the temperature at which the vapour pressure is equal to atmospheric pressure. We know that on the addition of a non-volatile liquid to a pure solvent, the vapour pressure of a solution decrease. Therefore to make vapour pressure equal to atmospheric pressure we have to increase the temperature of the solution. The difference in the boiling point of the solution and the boiling point of the pure solvent is termed as elevation in boiling point.

If T0b is the boiling point of the pure solvent and Tb is the boiling point of the solution then elevation in boiling point is given as

∆Tb =T0b-Tb

Experimental results show that there is a relation between elevation in boiling point and molality ‘m’ of the solute present in solution

∆Tb ∝ m

∆Tb = kb m

Where,

kb = molal elevation constant

Substituting the value of ‘m’ in the above relation we get

∆Tb = 1000 x kb x m2 / M2 x m1

Where,

m2 = mass of solvent in g

M1 = mass of solvent in kg

M2 = molar mass of solute

### Depression in Freezing Point

The freezing point of a substance is defined as the temperature at which the vapour pressure of its liquid is equal to the vapour of the corresponding solid. 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 a lower temperature. The difference between the freezing point of the pure solvent and its solution is called depression in freezing point.

If T0f is the boiling point of the pure solvent and Tf is the boiling point of the solution then depression in freezing point is given as

∆Tf =T0f-Tf

Just like elevation in boiling point, depression in freezing point is also directly related to molality ‘m’.

∆Tf = 1000 x kf x m2 / M2 x m1

Where,

k f = molal depression constant

m2 = mass of solvent in g

M1 = mass of solvent in kg

M2 = molar mass of solute

### Osmotic Pressure

When a semipermeable membrane is placed between a solution and solvent, it is observed that solvent molecules enter the solution through the semipermeable membrane and the volume of the solution increases. The semi-permeable membrane allows only solvent molecules to pass through it but prevents the passage of bigger molecules like solute. This phenomenon of the spontaneous flow of solvent molecules through a semipermeable membrane from a pure solvent to a solution or from a dilute to a concentrated solution is called osmosis.

The flow of solvent molecules through the semipermeable membrane can be stopped if some extra pressure is applied from the solution side. This pressure that just stops the flow of solvent is called osmotic pressure of the solution. Osmotic pressure is a colligative property as it depends on the number of solute present and not on the nature of the solute. Experimentally it was proved that osmotic pressure (⫪) is directly proportional to molarity(C) and temperature(T).

Mathematically, ℼ = CRT where R is the gas constant.

= (n2/V) RT

Here, V is the volume of solution in litres and n2 is moles of solute

If m2 is the weight of solute and M2 molar mass of solute, then n2= m2/M2

= W2 RT / M2V

Thus by knowing the values of ,w2, T and V we can calculate the molar mass of the solute.

Different Solutions

• Isotonic solution: Two solutions having the same osmotic pressure at a given temperature are known as an isotonic solution. When such solutions are separated by a semi-permeable membrane than there is no osmosis.
• Hypotonic solution: A hypotonic solution has a lower osmotic pressure than that of the surrounding i.e, the concentration of solute particles is less than that of the surrounding. If the hypotonic solution is separated by semipermeable membrane then water moves out of the hypotonic solution.
• Hypertonic solution: A hypertonic solution has a higher osmotic pressure than that of the surrounding i.e, the concentration of solute particles is more than that of the surrounding. If the hypertonic solution is separated by semipermeable membrane then water moves inside the hypertonic solution. For determining the molar mass, Osmotic pressure method has the advantage over other methods as pressure measurement is around room temperature. It is particularly useful for determination of the molar mass of biomolecules as they are unstable at higher temperatures.

## Van’t Hoff Factor

When the solute undergoes dissociation or association in solution, the number of particles in solution increases or decreases and thus, colligative properties changes accordingly. The extent of dissociation or association of the solute in a solution can be expressed by a factor called Van’t Hoff. ## Solved Problems

1. The molal elevation constant for water is 0.513o C kg mol. When 0.2mole of sugar is dissolved in 250g of water, calculate the temperature at which the solution boils under atmospheric pressure.

Solution:

The elevation in boiling point can be written as

∆Tb = moles of sugar x 1000 / weight of water in gram

∆Tb = 0.2 x 1000 / 250

∆Tb = 0.8

⇒ T0b-Tb = 0.8

For pure water, T0b =100oC

⇒ Tb= 0.8 + 100

=100.80 0C

2. A solution of CaCl2 was prepared by dissolving 0.0169g in 1 Kg of distilled water in ( molar mass of Ca2+= 41g mol and Cl= 35.5gmol). The molal depression constant of water is 2kg mol. The depression in freezing point of the solution is:

Solution:

Van’t Hoff factor(i) of CaCl2 is 3

Kf = 2 kg mol

Given,

Mass of CaCl2 (m2) = 0.0169g

Molar mass of CaCl2( M2) =112g

Weight of water(m1) =1000g

∆Tf = i x 1000X Kf x m2 / M2 x m1

= 3 x 1000 x 2 x 0.0169 / 112 x 1000g

= 9× 10-4

3. Hexane and heptane were mixed to form an ideal solution. At 1000C, the vapour pressure of two liquids( Hexane and heptane) are 190kPa and 45kPa respectively. What will be the vapour pressure of the solution obtained by mixing 25g of hexane and 35 g of heptane will be:

Solution :

No of moles of hexane,n1= 25/86= 0.29

No of moles of heptane, n2=35/100=0.35

𝜒1 = n1 / n1 + n2

𝜒1 = 0.29 / 0.29 +0.35

𝜒1 = 0.45

𝜒2 = 1 – 0.45

𝜒2 = 0.55

P = P01𝜒1 +P02𝜒2

= 1900.45 + 450.55

=110 kPa

4. On adding non-volatile solute to a solvent which of the following effect is observed;

a. Increase in vapour pressure

b. Increase in freezing point

c. Decrease in osmotic pressure

d. Decrease in vapour pressure

Solution: Option d

In a pure solvent, the entire surface is occupied by the molecules of the solvent. If a non- volatile solute is added to the solvent, the surface now has both solute and solvent molecules; thereby fraction of surface covered by solvent molecules gets reduced. Since the vapour pressure of the solution is solely due to solvent alone, at the same temperature the vapour pressure of the solution is found to be lower than that of the pure solvent.

5. 300 cm3 of an aqueous solution contains 1.56g of a polymer. The osmotic pressure of such solution at 270oC is found to be 2.57 ✕ 10-3 bar. Calculate the molar mass of the polymer

Solution:

Weight of polymer( W2) = 1.56g

Osmotic pressure (⫪) = 2.57 ✕ 10-3 bar

volume (V) = 300 cm3 = 0.3L

M2 = W2 RT / V

= 1.56 x 0.083 x 300 / 0.3 x 2.57 x 10-3

= 50381 g mol