Colligative Properties
Colligative properties refer to the physical changes that result from adding solute to a solvent. Colligative properties depend on the number of solute particles and the amount of solvent. Precisely, they depend on the ratio of the number of solute particles to the number of solvent particles in a solution. The number ratio can be expressed in the solution’s concentration units, such as molarity, molality, and normality. [1-4]
Colligative properties do not depend on the type of solute particles, although they do depend on the type of solvent. An assumption for colligative properties is that the solution must be ideal, and its thermodynamic properties must be analogous to those of an ideal gas.
Types of Colligative Properties
The four colligative properties are: [1-4]
Example
The following instances demonstrate colligative properties in solutions. When a small amount of salt is added to a glass of water, the solution’s freezing point significantly decreases compared to the water’s freezing point. Similarly, the solution’s boiling point rises, resulting in reduced vapor pressure. Moreover, alterations in the solution’s osmotic pressure also become evident.
Let us study briefly the abovementioned colligative properties.
1. Vapor Pressure Lowering
Every pure liquid exhibits a unique vapor pressure when in equilibrium with its liquid phase. The partial pressure associated with this equilibrium is intricately linked to temperature. In the case of solutions, their vapor pressure is notably lower than that of the pure solvent. The extent of this reduction depends upon the proportion of solute particles present, assuming the solute itself does not possess a substantial vapor pressure (nonvolatile).
The following formula gives the actual vapor pressure of a solution:
Psolution = χsolvent x Psolvent
Where:
– Psolution: Vapor pressure of the solution
– χsolvent: Mole fraction of the solvent
– Psolvent: Vapor pressure of the pure solvent
This equation is commonly referred to as Raoult’s law. The rationale behind vapor pressure depression is based on the assumption that solute particles occupy positions on the surface, displacing solvent particles. Consequently, this displacement hinders solvent evaporation.
2. Boiling Point Elevation
Due to the presence of a nonvolatile solute, the solution’s vapor pressure is lower than that of the pure solvent. A liquid boils when its vapor pressure becomes equal to the atmospheric pressure, which is 1 atm at sea level. To achieve this pressure, the solution must be at a higher temperature than the pure solvent’s boiling point.
The change in boiling point, denoted as ΔTb, can be easily computed using the formula:
ΔTb = i x Kb x m
Where:
– m is the molality
– Kb is the boiling point elevation constant
– i is the van’t Hoff factor
3. Freezing Point Depression
We have seen that the solution’s boiling point exceeds that of the pure solvent. However, an inverse effect is observed regarding the freezing point: The freezing point of the solution reduces below that of the pure solvent. This phenomenon can be visualized by considering how solute particles disrupt the cohesion of solvent particles during the formation of a solid, necessitating a lower temperature to induce solidification.
The change in freezing point (ΔTf) for a solution is given by:
ΔTf = i x Kf x m
Where:
– m is the molality
– Kf is the freezing point depression constant
– i is the van’t Hoff factor
4. Osmotic Pressure
A semipermeable membrane is a physical object that allows certain substances to pass through and prevents others. Placing it between a solution and solvent reveals an intriguing phenomenon: Solvent molecules permeate through the membrane into the solution, increasing its volume.
The semipermeable nature of the membrane allows passage solely for solvent molecules while effectively blocking larger solute molecules. This spontaneous movement of solvent molecules through the membrane, from a pure solvent to a solution or from a dilute to a concentrated solution, is termed osmosis.
Applying additional pressure from the solution side can obstruct the movement of solvent molecules across the semipermeable membrane. This defensive pressure against the solvent flow is the solution’s osmotic pressure.
Experimental evidence validates that the osmotic pressure (Π) is directly proportional to molarity (M), temperature (T), and van’t Hoff factor (i). Mathematically, this relationship can be expressed as
Π = i x M x R x T
where the gas constant is denoted as R.
This equation can also be represented as
Π = (n2/V) x R x T
where V signifies the volume of the solution in liters, and n2 represents the number of moles of the solute.
If the mass of the solute is denoted as m2 and its molar mass as M2, then n2 can be calculated as
n2 = m2/M2
It leads to the following formulation:
Π = (m2 x R x T)/(M2 x V)
Given the values of osmotic pressure (Π), solute mass (m2), temperature (T), and volume (V), it becomes feasible to calculate the molar mass of the solute through this equation.
Example Problems
Problem 1. Calculate the vapor pressure of a solution prepared by dissolving 25 g of glucose in 200 g of water at 27 °C. The vapor pressure of pure water at 27 °C is 26.74 torr. (Molecular masses of glucose and water are 180 g/mol and 18 g/mol, respectively)
Solution
Given m2 = 25 g, m1 = 200 g, T = 27 °C = 27 + 273 = 300 K, Psolvent = 26.74 torr
We use the following formula:
Psolution = χsolvent x Psolvent
To calculate the molar fraction of the solvent, we use the following formula:
χsolvent = n1/(n1 + n2)
Let us calculate the number of moles of glucose and water present in the solution.
n1 = 200 g/18 g/mol = 11.1 mol
n2 = 25 g/180 g/mol = 0.139 mol
We have,
χsolvent = 11.1 mol/(11.1 mol + 0.139 mol) = 0.988
And,
Psolution = 0.988 × 26.74 torr = 26.41 torr
Therefore, water’s vapor pressure is reduced from 26.74 torr to 26.41 torr.
Problem 2. A sample of 1 g of a nonvolatile organic compound is dissolved in 50 g benzene. The BP of the solution is 81.06 °C. BP of pure benzene is 80.08 °C. What is the molar mass of the solute? (The boiling point constant for benzene is 2.53 °C kg mol-1)
Solution
Given m2 = 1 g, m1 = 50 g, TB, solution = 81.06 °C, TB, solvent = 80.08 °C, and KB = 2.53 °C kg mol-1
The van’t Hoff factor of a nonvolatile solute is 1.
We use the following formula:
ΔTb = i x Kb x m
=> 81.06 °C – 80.08 °C = 1 x 2.53 °C kg mol-1 x m
=> m = 0.387 mol kg-1
The number of moles of the solute is
n = m x m1 = 0.387 mol kg-1 x 50 g x 10-3 kg/g = 0.019 mol
The molar mass of the solute is
M = m2/n = 1 g / 0.019 mol = 51.6 g/mol
