Colligative Properties

Colligative properties are properties of solutions that depend only on the number of solute particles present in the solution, not on their chemical nature.

Colligative properties change when you add more solute particles to a solvent, regardless of what the solute actually is.

Types of Colligative Properties

There are four types of colligative properties exhibited by the solution, which are,

1. Relative Lowering Of Vapour Pressure

The lowering of vapour pressure is the ratio of the vapour pressure of the solution to the vapour pressure of the pure solvent. If we add any non-volatile solute to the solvent, we can easily lower the vapour pressure of the solution. The image added below shows the relative lowering of the vapour pressure of any solution.

Suppose if P_o is the vapour pressure of the pure solvent and P_s is the vapour pressure of the solution, then the relative lower of the vapour pressure is calculated by

Relative Lowering of Vapour Pressure = (P_o - P_s)/P_o

The relative lowering of the vapour pressure of a solution having non-volatile solutes is equal to the mole fraction of the solute (X_Solute). If n is the number of moles of the solute and N is the number of moles of the solvent. Then we can say that,

(P_o - P_s)/P_o = n/(n + N) = X_Solute

2. Elevation in Boiling Point (ΔT_b)

The boiling point of the liquid is the temperature at which the vapour pressure of the liquid is equal to the atmospheric pressure. If we add any solute to the solvent, the boiling point of the solution is greater than the boiling point of the solvent.

Elevation of the boiling point is the difference between the boiling point of the solution and the boiling point of the solvent. If T_b is the boiling point of the solution, and T⁰_b is the boiling point of the pure solvent. Then the elevation of the boiling point is given as,

ΔT_b = T_b -T⁰_b

Experimentally, it is proven that ΔT_b is proportional to the molality (m) of the solute, then

ΔT_b ∝ m

ΔT_b = k_b m

Where k_b is the molar elevation constant.

ΔT_b = 1000k_b(m₂)/(M₂m₁)

where,

• m₂ is the mass of solute in g
• m₁ is the mass of solvent in g
• M₂ is the molar mass of the solute

3. Depression in Freezing Point (ΔT_f)

Freezing point of the liquid is the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase. If we add any solute to the solvent, the freezing point of the solution is lower than the freezing point of the solvent.

Depression of freezing point is the difference between the freezing point of the solution and the freezing point of the solvent. If T_f is the freezing point of the solution, and T⁰_f is the freezing point of the pure solvent. Then the depression in freezing point is given as,

ΔT_f = T⁰_f - T_f 

Experimentally, it is proven that ΔT_f is proportional to the molality (m) of the solute, then

ΔT_f ∝ m

ΔT_f = k_f m

Where, k_f is the molal depression (cryoscopic) constant.

ΔT_f = 1000k_f (m₂)/(M₂m₁)

where,

• m₂ is the mass of solute in g
• m₁ is the mass of solvent in g
• M₂ is the molar mass of the solute

4. Osmotic Pressure (π)

Osmotic Pressure is the colligative property of the solution and is defined as the difference needed to stop the flow of solvent across the semipermeable membrane.

We can also define the Osmotic Pressure as the minimum pressure that needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. It is also defined as the measure of the tendency of a solution to take in a pure solvent by osmosis. 

Osmotic Pressure is calculated by the formula,

π = CRT

Where,

• C is the Molar Concentration (molarity) of the solution
• R is Universal Gas Constant
• T is temperature

If the amount of solute is given in mass form, then:

π = w₂RT/M₂V

Where:

• w₂= mass of solute
• M_2​ = molar mass of solute
• V = volume of solution (in litres)

Types of Solutions

Different Types of solutions are given below:

Colligative Properties Examples

Various examples of the colligative properties can be easily seen in our daily lives. Such as when we add some salt to the water, the freezing point of the water increases, i.e. it freezes far below its normal freezing point. Also, the boiling point of the water increases if we add salt or sugar to the water. Similarly, adding alcohol to water decreases the freezing point of the water.

As these colligative properties are dependent on the concentration of the solution, let's first learn about the measure of concentration of the solution.

Molarity (M)

Molarity is the concentration of a solution, measured as the number of moles of solute per litre of solution. Molarity of any solution is calculated by the formula,

Molarity = Number of Moles of Solute/Volume of Solution in L

Molality (m)

Molality is the amount of a substance dissolved in a certain mass of solvent. It is defined as the number of moles of a solute per kilogram of a solvent. Molality of any solution is calculated by the formula,

Molality = Number of Moles of Solute/Weight of Solvent in Kg

Mole Fraction (x)

Mole fraction of a solute in a solution gives the ratio of the number of moles of the solute present in the solution to the total number of moles of the solute and the solvent present in the solution. There is another definition for mole fraction that is, "Mole fraction of a compound is the ratio of the number of moles of the compound to the total moles of compounds in the mixture." Mole fraction of any solution is calculated by the formula,

Mole Fraction = (Number of Moles of Solute)/(Total Number of Moles of Solute and Solvent)

Von’t Hoff Factor

For a solute that dissociates in the solution, its properties change by a factor and this factor is called the Von't Hoff Factor. Von't Hoff Factor is denoted using (i). It is calculated as,

i = (Observed Colligative Properties)/(Theoretical Colligative Properties)

or

i = (No. of Particles after Dissociation Or Association)/(No. of Particles is case of no Dissociation Or Association)

Related Articles:

• Effects of Change of Pressure
• Vapourization
• Partial Pressure Formula

Solved Examples on Colligative Properties

Question 1. Find the relative lowering of vapour pressure if 18 g of glucose is dissolved in 90 g of water.

Solution:

Given,

• Mass of glucose = 18 g
• Mass of water = 90 g

Molar Mass of Glucose = 180 g/mol

Number of moles of glucose(n_B) = 18/180 = 0.1 

Molar Mass of Glucose = 18 g/mol

Number of moles of water(n_A) = 90/18 = 5 

Relative lowering ΔP/P_A° is equal to X_B

X_B = n_B/(n_A + n_B)

X_B = 0.1/0.1 + 5 = 1/51

Relative Lowering of Vapour Pressure(ΔP/P_A°) = 1/51

Question 2. The boiling point elevation of a solution containing sucrose and water is 0.256 °C. The molal elevation constant of water is 0.512 °C/m, and the molar mass of sucrose is 342 g/mol. What is molality?

Solution:

Given:

• Boiling point elevation (ΔT_b) = 0.256 °C
• Molal elevation constant (K_b) = 0.512 

(ΔT_b) = K_b × m

0.256 = 0.512 × m

m = 0.5 mole/kg

Question 3. Calculate the freezing point depression and the freezing point after adding 100.0 g of table salt to 400.0 g of water. (K_f of water = 1.86) 

Solution:

Moles of NaCl = mass/molar mass

Moles of NaCl = 100.0/58.443 = 1.71107 mol

Mass of water = 400.0 g = 0.400 kg 

Molalilty(m) = (moles of NaCl)/(mass of water in kg)

m = 1.71107/0.400 = 4.2777

NaCl ⇢  Na⁺ + Cl^-

Van't Hoff factor(i) = Number of moles after dissociation/number of moles before dissociation 

i = 2

Freezing point depression constant for water, K_f = 1.86

Freezing point depression = i × K_f × m = 2 × 1.86 × 4.2777 = 15.9 °C

Freezing point of solution = (freezing point of water - freezing point depression) = 0.0 - 15.9

Freezing point of solution = -15.9 °C

Question 4. If 6.8% w/v of cane sugar is isotonic with 1.52% w/v of Thiocarbamide, if the molecular weight of cane sugar is 342, find the molecular weight of Thiocarbamide?

Solution:

In an Isotonic solution,

 π₁ = π₂  = i₂C₂RT  

Now, i₁C₁ = i₂C₂ 

Sugar and thiocarbamide are non-electrolytes,

So i = 1

Thus, C₁ = C₂ 

% (W/V) per cent  is the number of grams of solute in 100 mL of solution

C = (Number of moles of solute)/(Volume of solution in L)

C₁ = 6.8 × 1000/342 × 100

C₂ = 1.52 × 1000/x × 100

 As C₁ = C₂  

6.8 × 1000/342 × 100 = 1.52 × 1000/x × 100

x = 76

Thus, the weight of Thiocarbamide is 76 g

Question 5. Osmotic pressure of a solution of glucose is 117.4 atm. Find the molarity of the solution at 298 K.

Solution:

π = iCRT    

(Glucose is a Non-Electrolyte, i = 1)

117.4 = C × 0.0821 × 298

C = 4.8 mole/l

Thus, the concentration of glucose is 4.8 moles/litre

Question 6. x grams of solute is dissolved in 500 grams of solvent if the sum of the elevation of boiling point and depression in freezing point for sucrose in water is 5, find its molality. (if K_b = 0.52 and K_f = 1.86) find x?

Solution:

ΔT_f + ΔT_b = 5

We know that,

• ΔT_f = i × k_f × m
• ΔT_b = i × k_b × m

Here, i = 1 (for a nonelectrolyte like sucrose)

k_f × m + k_b × m = 5 

2.38 × m = 5

m = 2.1 mole/kg 

Molality = Number of Moles of Solute/Weight of Solvent in Kg

(x × 1000)/(342 × 500) = 2.1 

x = 359 gram

Thus, the solute dissolved is 359 grams.